Chapter 6 Mechanics III

examples of nonconservative forces

friction, air resistance, viscous drag

Examples of conservative forces

gravitational, elastic, electric

An object of mass m is projected straight upward with an initial speed of v0 at time t = 0. Use Conservation of Total Mechanical Energy to find its maximum height.

h = v^2 / 2g

One month, your electric bill states that you used 500 kWh of electricity, at a cost of 8 cents per kWh. What is a kWh, and mow much is your electric bill that month?

kWh = kilowatt hour 500 kWh x .08 = $40

Momentum units

kg m/s

Collisions equation

m1v1 + m2v2 = m1v1' + m2v2'

If cosine theta = 90, work is

0

What magnitude of work must be done to bring a 1000-kg car, moving at 20 m/s, to rest? A. 1.8 × 10^6 J B. 1.0 × 10^5 J C. 2.0 × 10^5 J D. 4.0 × 10^5 J

2.0 × 10^5 J of work must be done to bring a 1000-kg car, moving at 20 m/s, to rest. Find the kinetic energy, and do an amount of work equal to the negative of that value (the work-energy theorem). Since KE = 1/2 mv2 = 1/2 (1000)(20)2 = 2 × 10^5 J, we must do -2 × 10^5 J of work to stop it. Since the question asks for the magnitude, the answer is 2.0 × 105 J.

A tension force in a rope of 200N is directed 45 degrees above the horizontal pulls a 50kg crate to the right for 5m. If uK = .2, find the work done by each of the following forces acting on the object and the net work done by the object

340J

Which of the following expressions is equal to a watt? A. (kg)(m)2/(sec)3 B. (kg)(m)/(sec)3 C. (kg)2(m)2/(sec)2 D. (kg)(m)/(sec)

A

Which of the following situations requires the greatest power? A. 200 J of work done in 10 minutes B. 10 J of work done in 20 minutes C. 20 J of work done in 10 minutes D. 100 J of work done in 20 minutes

A

In this situation, assume the mass of the crate, m , is 20 kg and the coefficient of kinetic friction between the crate and the floor is .4. If F = 100 N and d = 6m, A. How much work is done by F? B. How much work is done by the normal force? C. How much work is done by gravity? D. How much work is done by the force of friction E. What is the total work done on the crate?

A. W = F*d * cos theta W = 100N * 6 = 600J B. 0, because the normal force is perpendicular to the floor C. 0, because cos 90 = 0 D. Ff = uKFN Ff = (.4)(200) Ff = 80N W = F *d * cos theta W = 80N * 6m * -1 (because cos 180 = -1) W = -480J E. W= 600 J - 480J W = 120J

You are lifting bricks, each with a mass of 2 kg from the floor up to a shelf that is 1.5 m high. A. How much work do you need to perform lifting each brick? B. If you can place 20 bricks on the shelf every minute, what is your power output? C. If you continue this effort for an hour, how many calories of work will you do? 1 cal = 4184J

A. W = Fd W = (20N)(1.5) W = 30 J B. P = W/t P = (30J)(20 bricks) / (60s) P = 10 W C. W = Pt W = 10W * 3600 s W = 36,000 J W = 36,000 J * (1 cal / 4184J) = 9 calories

An object of mass 10kg whose initial speed is 4 m/s is accelerated until it achieves a final speed of 9 m/s. A. How much work was done on the object? B. If the acceleration took place over a displacement f of magnitude 13m, and the force F exerted on it was constant and parallel to d, what was F?

A. W = delta KE W= KE(f) - KE(i) W = 1/2mv^2 - 1/2mv^2 W = (.5)(10)(81) - (.5)(10)(16) W = 405J - 80J W = 325J B. W = Fd F = W/d F = 325J / 13m F = 25N

A bobsled team of four men, each of mass m, pushes the sled of mass M over a horizontal displacement d before leaping into the sled and beginning to slide down the hill. The total length of the winding track from the top of the hill to the bottom is L, and the height difference between the top of the hill and the bottom is Δh. If the sled starts at rest and we assume the whole track is frictionless, how does the final kinetic energy of the sled at the bottom of the hill (prior to the brakes being applied) compare to the initial potential energy of the sled at the top of the hill? A. KE f > PEi B. KE f = PEi C. KE f < PEi D. It is not possible to uniquely determine the relationship between these two energies.

A;

A object is raised to a height of 16 m and released from rest. At the instant that the object is 12 m above the ground, what fraction of its total mechanical energy is in the form of kinetic energy? (Ignore air resistance; assume that potential energy is measured relative to the ground.) A. 1/4 B. 3/8 C. 1/2 D. 3/4

A;

An increase in which of the following properties of a projectile will NOT increase its kinetic energy at a given instant in time? A. Its height B. Its momentum C. Its velocity D. Its mass

A;

As a 5-kg object travels down a ramp, gravity does 60 J of work and friction does -20 J. If the object started from rest, what is its final speed? A. 4 m/s B. 1 m/s C. 8 m/s D. 2 m/s

A;

A 2-kg block slides down a 3-meter-long, frictionless 30° incline. How much work does gravity do on the block? A. 30 J B. 60 J C. 40 J D. 50 J

A; W = mgd W = (2)(10 sin 30)(3) W= (2)(10)(3)(.5) W = 30J

A 10-kg object moves from Position #1 to Position #2 close to the surface of the Earth. In so doing, its gravitational potential energy decreases by 200 J. How much work was done by the gravitational force on this object as it moved from Position #1 to Position #2? A. 200 J B. -200 J C. -100 J D. 100 J

A; A 10-kg object moves from Position #1 to Position #2 close to the surface of the Earth. In so doing, its gravitational potential energy decreases by 200 J. 200 J was done by the gravitational force on this object as it moved from Position #1 to Position #2. Work done by the gravitational force equals the opposite change in gravitational potential energy. They have the same magnitude, but opposite sign. When an object is falling, its displacement vector points in the same direction as the force of gravity, thus gravity does positive work on the object, while the object loses gravitational potential energy. Work done by gravity will always have the opposite sign of the change in gravitational potential energy. If the work is positive, the object fell and lost potential energy. If the work is negative, the object was raised and gained potential energy.

An electric crane hoists an object weighing 4000 N to the top of a building. The crane raises the object straight upward at a constant rate. If it takes 60 seconds to lift the mass 300 m, at what rate is energy consumed by the electric motor in the crane? (Note: Ignore all forces of friction.) A. 20.0 kW B. 2.0 kW C. 0.8 kW D. 10.0 kW

A; An electric crane hoists an object weighing 4000 N to the top of a building. The crane raises the object straight upward at a constant rate. If it takes 60 seconds to lift the mass 300 m, energy consumed at 20.0 kW by the electric motor in the crane. Since the object's velocity is upward and constant and the force necessary to pull it upward is also upward and constant, we may use the equation P = Fv. Since the object moves 300 meters in 60 sec, its velocity is v = (300 m)/(60 s) = 5 m/s. Thus, the rate at which energy is used is P = (4000 N)(5 m/s) = 20,000 W = 20 kW.

As a crate slides down from the top of a 2-meter-long inclined plane, the coefficient of friction is 0.4. Calculate the work done by friction if the angle of incline is 30° and the mass of the crate is 10 kg. A. -68 J B. -20 J C. -39 J D. -34 J

A; As a crate slides down from the top of a 2-meter-long inclined plane, the coefficient of friction is 0.4. The work done by friction if the angle of incline is 30° and the mass of the crate is 10 kg is -68 J. The strength of the friction force is equal to the coefficient of friction times the magnitude of the normal force (which is mg cos θ), and since the friction force is opposite in direction from the displacement ∆s, the work done by friction will be negative:

By applying a large horizontal force, a man pushes a heavy crate along a horizontal floor. While he pushes the crate a distance d, the frictional force does -W1 joules of work. A small child then sits on top of the crate, and the man pushes the crate (and child) a distance d. If the frictional force does -W2 joules of work during this second displacement, then which of the following is true? A. W1 < W2 B. W1 > W2 C. Cannot be determined from the information given D. W1 = W2

A; By applying a large horizontal force, a man pushes a heavy crate along a horizontal floor. While he pushes the crate a distance d, the frictional force does -W1 joules of work. A small child then sits on top of the crate, and the man pushes the crate (and child) a distance d. If the frictional force does -W2 joules of work during this second displacement, then W1 < W2 is the true relation. With the child on top of the crate, the normal force exerted by the floor on the crate must increase (to support the additional weight). Since N increases, the force of kinetic friction increases, so the magnitude of the work done by friction during the second displacement is greater: W2 > W1.

Late for a flight, a man runs through the airport concourse at 6 m/s pulling his rolling 15 kg suitcase with a (nearly) horizontal tension force of 200 N. How much power is the man exerting in pulling the suitcase during this process? A. 1200 W B. 120 W C. 80 W D. 0 W

A; For a constant force parallel to velocity, power is given by P = F ? v. In this case that yields P = 200 N × 6 m/s = 1200 W.

A pulley system is used in an auto repair shop to lift a 200 kg engine block 1.5 m out of a car. The lower pulley is attached directly to the engine block, and there are 8 strands of rope attaching the lower pulley to the upper pulley. The repairman pulls through 12 m of rope in the process of lifting the engine block. How much work is done during this process? A. 3000 J B. 24 kJ C. 300 J D. 375 J

A; One critical fact to remember about the mechanical advantage supplied by simple machines (like systems of pulleys) is that while they reduce the force necessary to lift a mass, they do not change the work done in the process. Therefore, W = mgh = 200 kg × 10 m/s2 × 1.5 m = 3000 J. In this case, the repairman must pull eight times the length of rope, applying one eighth the force, to lift the engine block as he would if he simply tied a rope directly to the block and lifted it straight up 1.5 m.

A man decides to move into an apartment two floors above where he currently lives. The elevator is broken, so he needs to move all of his possessions up the stairs to the new apartment. The man has the choice between doing all the moving himself or hiring two professional movers to do the job, each of whom can move twice as many kilograms of material per hour as the man can. Presuming the man and the movers are perfectly capable of moving all the belongings safely to the new apartment, what is the primary physical difference between the two moving choices? A. The movers together will exert more power than the man will in moving his belongings because they will work faster. B. The man will exert more power than both of the movers together because it will be twice as hard for him to complete the move. C. The movers together will do more work than the man will because they will complete the task sooner. D. The man will do more work than both of the movers together because it will be twice as hard for him to complete the move.

A; The amount of work done will be the same in either case because ultimately the same amount of mass will be moved upward the same displacement against gravity, so choices C and D are eliminated. The scenario informs us that the movers each work twice as fast as the man, and there are two of them to one of him, so they will take one fourth as long to finish the task as the man. Power is work / time, so less time with the same amount of work means that the movers will exert more power during the move.

Two pulleys--one mounted in the ceiling, another anchored to a mass Msuspended above the ground below--have a rope looped over them three complete times, so that there are six strands of rope running between the two pulleys. One end of the rope is tied to the center of the top pulley, the other is being held by a man standing next to the mass. The man pulls down with a tension T on that strand of rope causing the mass to rise at a constant speed. What is the net force pulling up on the bottom pulley? A. Mg B. Mg / 3 C. Mg / 6 D. Mg - T

A; This is an easy problem to over-think, so consider the equilibrium of the system. The net force acting on the mass M must be zero, so the total tension force up for all the strands together is Mg. Note that the question is not asking for the force of tension.

A slow-moving conveyor belt system is used in a mining operation to transport tons of ore at a constant speed from deep below the earth to the surface, where it is loaded into large gasoline powered trucks and transported rapidly along a level horizontal road to processing facilities. The conveyor belt is driven by a large gasoline-powered motor. What best describes the work and energy of the system? A. Gasoline's chemical potential energy is transformed into mechanical work done on the ore to increase first its gravitational potential energy and then its kinetic energy, as well as into thermal energy within the system. B. The ore's gravitational potential energy is transformed into its kinetic energy. C. Gasoline's chemical potential energy is transformed into the ore's kinetic energy, which does work on the ore to give it gravitational potential energy and thermal energy. D. The heat of the liquid gasoline is transformed into the gravitational potential energy and kinetic energy of the ore.

A; Work is done on the ore first to lift the ore (increasing its gravitational potential energy) and then to get it moving on the trucks (increasing its kinetic energy). This eliminates choice B, which would be the correct answer for a case such as the ore rolling downhill, as well as choice C, which has the wrong order for the potential and kinetic energies. Both the conveyor belt and the trucks are run by gasoline powered engines, so the work done on the system comes from the transformation of the chemical potential energy in the fuel into the two different forms of mechanical energy and into thermal energy, i.e., heat, as the engines heat up. Note that the heat of the gasoline comes from combustion, which converts the liquid into a gas, eliminating choice D.

A 1000 kg car moving at 20 m/s up a slight incline brakes to a stop. If the braking force has a magnitude of 1500 N and the car goes up 5 m during the braking process, how far did the car move during that time? A. 1000 m B. 100 m C. 50 m D. 10 m

B

A person's power expenditure is being monitored. If the amount of work is doubled and the time required to complete it is halved, then the power output: A. decreases by a factor of 4. B. increases by a factor of 4. C. remains constant. D. increases by a factor of 2.

B

Which of the following are the dimensions of a watt? A. [M][L] / [T]2 B. [M][L]2 / [T]3 C. [M]2[L]2 / [T]2 D. [M][L]3 / [T]2

B

A lever is used to lift a 10 kg mass to a height of 2 m. If the person on the other side of the lever exerted a force of 50 N over a vertical distance of 5 m to accomplish this lift, what is the efficiency of the lever system? A. 125% B. 80% C. 12.5% D. 8%

B;

The mechanical advantage of a machine is a ratio of the force required to lift an object a given distance to the force required to move that object the same vertical distance using a simple machine. A ramp is one example of a simple machine that allows one to exert less force in moving an object from a lower to a higher point. Suppose a box is pushed along a 5 meter long ramp with a rise of 2 meters (with negligible friction between the box and the ramp). What mechanical advantage is gained by using the ramp instead of just lifting the box straight up? A. 5 B. 2.5 C. 0.2 D. no advantage

B; Because the work involved is the same whether the mass is lifted straight up or pushed up the ramp, mechanical advantage can also be calculated as the ratio of the distance over which the force is exerted using the simple machine to the vertical displacement: W = Fd = mgh MA = mg/F(applied) MA = d/h MA = 5/2 = 2.5 Can also use F(machine) / F(effort) 5/2 = 2.5

A 10-kg object is projected straight upward with an initial kinetic energy of 1000 J. How high will it go above its launch point? (Ignore air resistance.) A. 5 m B. 10 m C. 50 m D. 20 m

B; A 10-kg object is projected straight upward with an initial kinetic energy of 1000 J. It will go 10 mabove its launch point. (Ignoring air resistance.) This is kinetic energy that's being converted (entirely) into potential (at its highest point), rather than the other way around.

A car weighing 8500 N and traveling 20 m/s engages its brakes. The car skids along the pavement for 200 m before coming to rest. What is the coefficient of friction between the road and the car's tires? A. 0.5 B. 0.1 C. 0.8 D. 0.2

B; A car weighing 8500 N and traveling 20 m/s engages its brakes. The car skids along the pavement for 200 m before coming to rest. The coefficient of friction between the road and the car's tires is 0.1. The work-energy theorem guarantees that the total work done on the car is equal to its change in kinetic energy. If the strength of the friction force is denoted by Ff , then Ff=µ N =µ mg, so the work done by friction as the car covers a displacement of magnitude d is -Ffd = -µ mgd. Setting this equal to - 1/2 mv2, which is the car's change in kinetic energy, we find that

A stone of mass m is dropped from a height h. If air resistance is negligible, which one of the following statements is true concerning the stone as it strikes the ground? A. Its speed is proportional to h. B. Its kinetic energy is proportional to h. C. Its kinetic energy is proportional to h2. D. Its speed is proportional to h2.

B; A stone of mass m is dropped from a height h. If air resistance is negligible, as the stone strikes the ground, its kinetic energy is proportional to h. Since the only force doing work on the stone is gravity, a conservative force, the stone's energy is conserved. Its potential energy is wholly converted into kinetic energy. Therefore, we get mgh = KEf , so KEf ∝ h.

A person rides a bike uphill at a constant speed. What best describes the work and energy changes within the system? A. There is no work being done by the person because the speed isn't changing. B. The person is doing external work on the system that increases the gravitational potential energy. C. The nonconservative work being done by kinetic friction on the bike tires cancels out the positive external work the person does in pumping the pedals. D. It is impossible to say how much work is being done without knowing the angle of inclination of the hill.

B; External work done on a system in which mechanical energy is conserved increases either the kinetic or the potential energy (or both) of the system. There are no nonconservative forces like kinetic friction (since the tires aren't slipping), so mechanical energy is conserved; with a constant speed, kinetic energy isn't changing, so the increase in potential energy for the uphill movement must be accomplished by the work done on the system by the person peddling the bike.

A book (m = 1 kg) falls off of the top shelf (h = 1.5 m) of a bookcase and hits the floor. It is then accidentally kicked across the floor, moving a distance of 3 m before friction (µk = 0.25) brings it to a stop. The book is then returned to the top shelf 2 m to the right of its original position. What is the change in the gravitational potential energy of the book during this process A. 20 J B. 0 J C. -15 J D. -22.5 J

B; Gravity is a conservative force, so the change in gravitational potential energy depends only upon the initial and final heights of the book, not on the path the book takes to get from one to the other: ΔPEgrav = mgΔh. Because there is no change in height in moving from the top shelf back to the top shelf, there is no change in energy.

An object of mass m is dropped from a height of h meters; its speed at impact with the ground is vm/s. If an object of mass 4m were dropped from a height of h meters, determine its speed at impact. (Ignore air resistance.) A. 4v B.v C. 2v D. 16v

B; How did the mass of the object enter into the equation for its impact speed, v? Answer: It didn't

In an avalanche, a 2000 kg boulder slides down a straight icy (frictionless) hill 1 km long. The acceleration of the boulder is 4 m/s2. How much work does gravity do on the boulder during this slide? A. 2 × 10^7 J B. 8 × 10^6 J C. 2 × 10^4 J D. 8 × 10^3 J

B; It is important not to over-think work problems: if the component of a force in the direction of displacement is known, then work is simply the product of that component and displacement. In this case, gravity causes the acceleration down the frictionless hill, so the component of the gravitational force is simply Fgrav down hill = ma = (2000)(4) = 8000 N. The work done then is Wgrav = (8000 N)(1000 m) = 8 × 106 J. A longer way to do this is to find the angle of the incline, knowing that acceleration = g sinθ. The angle can then be used to find the height h of the slope and the work done by gravity would be equal to mgh.

How much kinetic energy does a 1500 kg car have when it is moving at a speed of 20 m/s? A. 6 × 10^5 J B. 3 × 10^5 J C. 1.5 × 10^5 J D. 3 × 10^4 J

B; KE = 1/2mv^2 KE = (.5)(1500kg)(400m^2/s^2) KE = 3 x 10^5 J

A man drags a wooden crate (m = 50 kg) across the ground (µk = 0.2) with a rope at an angle of 45° up from the horizontal. If the tension in the rope is 200 N, how much work is done by the pulling force in moving the crate over a displacement of 5 m? A. 1000 J B. 700 J C. 340 J D. 200 J

B; The work done by a single force is product of the parallel components of that force and the displacement vector: W = Fd cosθ. Therefore WT = (200)(5)cos 45° ≈ (200)(5)(0.7) = 700 J.

A car of mass m drives around a circular banked track of radius R at constant speed v, and maintains a constant height h. What best describes the work done by the normal force on the car during one quarter of a revolution around the track? A. WN < 0 B. WN = 0 C. WN > 0 D. The work done by the normal force can be positive or negative depending upon the direction the car is moving.

B; Work done by a force on an object depends upon the component of the force along the direction of the displacement of the object. Anything moving in a circular path is, at any given instant, displacing in a direction tangent to the path (the instantaneous velocity), while it is accelerating along the radius toward the center of the circle (centripetal acceleration). Thus the force and displacement are perpendicular, and so the work done by the force is zero. It is also useful to consider that, because the car is not changing height or speed, it is neither gaining nor losing mechanical energy, so no work is being done on it by an outside force.

Drag is a force that acts on any object moving through a fluid. The drag force is always directed opposite the velocity of the object. Suppose a ball is thrown straight up into the air and a few seconds later it is caught. From the moment the ball leaves the thrower's hand to the instant before it is caught, what can be said definitively about the work done by drag? A. Wdrag > 0 B. Wdrag = 0 C. Wdrag < 0 D. The work done by drag over the flight is equal to the negative of the work done by gravity.

C

A mass m is released from rest at the top of a frictionless inclined plane with a 3 meter base and a 4 meter height. With what speed does the mass reach the ground at the base of the plane? A. 4 m/s B. 5 m/s C. 9 m/s D. 11 m/s

C;

Consider the pulley system below. Suppose there is nonzero kinetic friction(coefficient of kinetic friction µk) between the mass m and the tabletop on which it slides. A horizontal pulling force is applied to the left which causes the mass to move a displacement d at a constant speed. How much work is done by this pulling force? A. 0 J B. µkmgd C. (µkmg + Mg)d D. (Mg - µkmg)d

C;

Two cars of equal mass are traveling opposite directions on a highway. If one is moving twice as fast as the other, what is the relationship between their kinetic energies? A. One is a quarter of the value and the opposite sign of the other. B. One is half the value and the opposite sign of the other. C. One is a quarter the value and the same sign as the other. D. One is half the value and the same sign as the other.

C;

An automobile with a certain shape experiences a drag force due to air resistance that is, in Newton's, equal to 1/3 the square of the cars speed, in meters per second. How much power would the engine have to supply to the wheels to balance this drag force when the car is moving at a constant speed of 30 m/s. A. 10 W B. 300 W C. 9 kW D. 27 kW

C; P = Fv F(drag) = 1/3 v^2 P = 1/3v^2 *v P = 1/3 v^3 P = 1/3 (30*30*30) P = 1/3 (27,000) P = 9,000 W, or 9kW

A 1-kg ball is dropped from a height of 6 meters. As it falls, it is constantly acted upon by air resistance, whose average force on the ball is 3.3 N. Taking this into account, calculate the speed with which the ball hits the ground. A. 10.0 m/s B. 10.6 m/s C. 9.0 m/s D. 11.1 m/s

C; A 1-kg ball is dropped from a height of 6 meters. As it falls, it is constantly acted upon by air resistance, whose average force on the ball is 3.3 N. Taking this into account, the speed with which the ball hits the ground is 9.0 m/s. Use the conservation of energy equation (including the term for the work done by friction):

A 2000-kg airplane flying at 50 m/s is slowed by turbulence to 40 m/s over a distance of 150 m. How much work was done on the plane by the turbulent air? A. -100 kJ B. -1800 kJ C. -900 kJ D. -10 kJ

C; A 2000-kg airplane flying at 50 m/s is slowed by turbulence to 40 m/s over a distance of 150 m. -900 kJ of work was done on the plane by the turbulent air. Use the work-energy theorem. The work performed on the plane by the turbulent air is equal to the change in the plane's kinetic energy, which is ΔKE = 1/2 m(v^2 - v0^2) = (1000)(40^2-50^2) = -900 kJ. Note that turbulence as a nonconservative interaction (like friction) can only do negative work.

A crate with mass 50 kg is pushed across a horizontal floor at a constant speed of 1 m/s for 4 seconds by a horizontal force F of magnitude 100 N. How much work is done by F? A. 0 J B. 200 J C. 400 J D. 100 J

C; A crate with mass 50 kg is pushed across a horizontal floor at a constant speed of 1 m/s for 4 seconds by a horizontal force F of magnitude 100 N. 400 J is done by F. Work is force times parallel displacement, so we need to find the displacement through which the force acts, which is the distance the crate moves. If the crate's speed is a constant 1 m/s for 4 seconds, it must travel 4 m. This gives W = Fd = (100 N)(4 m) = 400 J

A laborer expends 800 J to lift a block to a height h. He then repeats the task using a simple non-motorized pulley system that reduces by half the input force he must provide. With the pulley system in operation, how much work must the laborer perform in order to lift the block to height h? A. 200 J B. 1600 J C. 800 J D. 400 J

C; A laborer expends 800 J to lift a block to a height h. He then repeats the task using a simple non-motorized pulley system that reduces by half the input force he must provide. With the pulley system in operation, the laborer must perform 800 J in order to lift the block to height h. Machines reduce the amount of force required to perform a task-at the expense of increasing the distance through which the force acts. They do not reduce the amount of work required; that is fixed. If 800 J of energy must be expended to lift the block to a certain height, then, even if a perfectly efficient pulley is used, the laborer must still do a total of 800 J of work to lift the block to the same height.

Adjustments are made to a machine that allow it to provide less energy at any given moment, but that allow it to operate for a greater length of time. The power provided by the machine has been: A. increased. B. changed in a manner that can't be predicted. Your Answer C. decreased. D. unchanged.

C; Adjustments are made to a machine that allow it to provide less energy at any given moment, but that allow it to operate for a greater length of time, therefore, the power provided by the machine has been decreased. By definition, power is the rate of energy transfer, or the energy transferred divided by time. Thus, less energy output over a greater length of time will certainly have the effect of decreasing the power of the machine.

An object weighing 100 N is traveling horizontally with respect to the surface of the earth in the absence of air resistance at a constant velocity of 5 m/s. What is the power required to maintain this motion? A. 20 W B. 200 W C. 0 W D. 500 W

C; An object weighing 100 N is traveling horizontally with respect to the surface of the earth in the absence of air resistance at a constant velocity of 5 m/s. 0 W is required to maintain this motion since there is no force component parallel to the direction of displacement, therefore no work is being done and no power is being exerted.

Example of when the work = 0

Centripetal force, normal force

A block with a mass of 5 kg starts at rest at the higher end of a 5 meter long tilted ramp. The difference in height between the high and low ends of the ramp is 3 meters. The ramp itself is smooth enough that friction can be ignored as the block slides down, but once the block reaches the ground, it experiences a kinetic frictional force as it slides to a stop. If the block slides to a stop in 6 meters on the horizontal ground, what is µk between the block and the ground? A. 0.83 B. 0.67 C. 0.5 D. 0.33

C; Kinetic friction does the nonconservative work to bring the sliding block to a stop after its initial gravitational potential energy has been transformed into kinetic energy. Starting from the total energy conservation equation, the value of the coefficient of kinetic friction can be found from the work expression. Note that friction works over the horizontal ground, so the normal force is just mg, and that the block begins and ends at rest, so ΔKE = 0:

Two blocks start at height h: block 1 on a frictionless ramp of constant angle of inclination, block 2 suspended in midair. Both blocks are released at the same moment. How does the average power exerted on the blocks during their descent compare? A. P 1 > P2 B. P 1 = P2 C. P 1 < P2 D. Cannot be determined based on the information given.

C; Power is work / time. The work done by gravity on each block is the same, because the change in height of each block is the same. The normal force on block 1 does no work because it is perpendicular to displacement. Since there is no friction on the plane, the net work done on each block must therefore be the same. The time it takes block 2 to descend over the displacement h is less than the time it takes block 1 to displace a longer distance down the ramp (h / sinθ), both because block 2 undergoes less displacement and because block 2 experiences greater acceleration (g versus g sinθ). Less time for the same work means more power.

A 2.5-kg mass is projected straight upward with an initial kinetic energy of 980 J. If air resistance is ignored, how much kinetic energy will this projectile have as it strikes the ground? A. 1470 J B. 1960 J C. 980 J D. 490 J

C; Since air resistance is to be ignored, the original 980 J of kinetic energy is converted into 980 J of potential energy at the top of its path, which is in turn converted back into 980 J of kinetic energy as it strikes the ground. (The value given for the mass of the object is irrelevant.)

A machine that consumes 500 watts takes 10 sec to lift an object with a mass of 1 kg from the ground to a high platform. Assume that the machine is perfectly efficient. The object is then pushed off the platform and falls freely to the ground. What is the speed of the object at the moment of impact? A. 10 m/s B. 32 m/s C. 100 m/s D. 50 m/s

C; Since work is equal to power times time, W = Pt, the amount of work done by the machine is W = (500 W)(10 s) = 5000 J. If the machine is perfectly efficient, then the object's gravitational potential energy was increased by 5000 J as a result of being lifted up to the platform. This potential energy will be converted to kinetic energy as the object strikes the ground. Thus, setting 1/2mv2 equal to 5000 J, we find that

As a crate (of mass m) slides down a frictionless incline (with angle θ), a constant horizontal force F, parallel to the base of the incline, is applied to the crate so that the crate's speed down the incline remains constant. F = mg tan θ If the vertical rise of the incline is h meters, determine the work done by the horizontal force F as the crate slides down the incline. A. -mgh sin θ cos θ B. -mgh sin2 θ C. -mgh D. -mgh cos2 θ

C; The component of the force directed up the ramp is F cos θ, and the distance along the ramp is h/(sin θ). Now, from the figure above, the direction of F cos θ (up the ramp) is opposite to the direction of the displacement (down the ramp). Therefore, the work done will be negative. Using the value of F, we find: W = -F cos theta *(h/sin theta) = -mgsin theta / cos theta * cos theta * h/sin theta = -mgh Alternatively, one can use the relationship between work and gravitational potential energy to solve this problem. Since the object is traveling at a constant speed in the same direction, the net force on the crate is zero, and thus the component of the force of gravity going down the plane is equal to the component of the applied force going up the incline. The force of gravity is doing positive work on the crate, but since the kinetic energy of the crate is not changing, the applied force is doing the same magnitude of work on the crate, but negative. Remember that the work done by gravity is equal to the opposite change in gravitational potential energy. The crate lost mgh joules of gravitational potential energy, meaning gravity did mgh joules of work. Therefore, the applied force did -mgh joules of work.

Two pulleys--one mounted in the ceiling, another anchored to a mass Msitting on a platform below--have a rope looped over them three complete times, so that there are six strands of rope running between the two pulleys. One end of the rope is tied to the ceiling adjacent to the top pulley, and the other is being held by a man standing next to the mass. The man pulls down with a tension T on that strand of rope. If the mass just barely lifts off the ground, what is true of T? A. T = Mg B. T = 6Mg C. T = Mg / 6 D. T = Mg / 3

C; The pulleys create a mechanical advantage by multiplying the tension force by the number of strands running up from the bottom pulley. A pulley is simply a means of changing the direction of the tension force: it does not change its magnitude, nor does it change the basic property of ropes or strings, namely that they cannot push. This means that every strand connected to the bottom pulley applies a pulling force upward of magnitude T, so 6T = Mg (the weight of the mass being the minimum applied force upward required to lift it). Thus T = Mg / 6.

A block slides down a frictionless ramp of height h. It reaches speed v at the bottom of the ramp. To reach speed 2v, the block would have to slide down a ramp of height: A. (1.4)h B. 2h C. 4h D. 6h

C; With no friction, mechanical energy is conserved. Thus we have . So if v doubles, hmust quadruple.

Three people of equal mass are in an apartment building. Person 1 climbs the stairs slowly from the first to the third floor. Person 2 runs (taking half as much time as person 1) up the stairs from the second to the fourth floor. Person 3 takes the elevator (taking half as much time as person 2) from the fourth to the sixth floor. Assuming the floors are evenly spaced, which best describes the relations among the changes in the gravitational potential energy of the three? A. 3ΔPE1 = 2ΔPE2 = ΔPE3 B. 4ΔPE1 = 2ΔPE2 = ΔPE3 C. ΔPE1 = ΔPE2 = ΔPE3 D. ΔPE1 = 2ΔPE2 = 4ΔPE3

C; ΔPEgrav = mgΔh and is not functionally dependent upon the time it takes to change height nor (for changes small relative to the radius of the earth) upon the differences among the initial heights of the people, only upon their changes in height. Each person goes up two floors, so each has the same Δh.

A 10-kg object is dropped from a height of 100 meters. How much gravitational potential energy has it lost when its speed is 30 m/s? (Ignore air resistance.) A. 7750 J B. 2250 J C. 5500 J D. 4500 J

D;

A typical gasoline engine in a car has an efficiency of about 20%. About how many joules of thermal energy released by the internal combustion of gasoline and air are needed to accelerate a 1000 kg car from rest to a speed of 20 m/s on a flat road? A. 4 × 10^4 J B. 2 × 10^5 J C. 4 × 10^5 J D. 1 × 10^6 J

D;

If the height of an object above the earth is doubled, its gravitational potential energy relative to the ground will be: A. quadrupled. B. halved. C. unchanged. D. doubled.

D;

It takes 2.0 × 10^5 J of work to bring a 1000-kg car, moving at 20 m/s, to rest. What is the average power needed to stop the same car in 4 seconds? A. 500 kW B. 5000 kW C. 5 kW D. 50 kW

D;

Using the ramp on the back of a moving van, a mover who can apply a maximum force of 400 N is just able to move a 2000 N piano onto the van. If the bed of the moving van is 1 meter off of the ground, how long is the ramp? A. 8 m B. 2 m C. 4 m D. 5 m

D;

How much work is done by the gravitational force as a 10-kg object is lifted from a height of 1 m above the ground to a height of 3 m above the ground? A. 300 J B. -300 J C. 200 J D. -200 J

D; -200 J is done by the gravitational force as a 10-kg object is lifted from a height of 1 m above the ground to a height of 3 m above the ground. Eliminate both positive choices: since the object's displacement is upward, but the force of gravity is downward, the work done by gravity on this object must be negative. The value is W = -mg∆h = -(10)(10)(3 - 1) = -200 J

A 10-kg mass is dropped from a height of 125 m. What is its speed at impact with the ground? (Ignore air resistance.) A. 75 m/s B. 20 m/s C. 125 m/s D. 50 m/s

D; 10-kg mass is dropped from a height of 125 m. Its speed at impact with the ground is 50 m/s.(Ignoring air resistance.) At the beginning, all its energy is in potential form; as it slams into the ground, all its energy has been converted into kinetic form. Therefore,

A 2-kg object is at a height of 10 m above the surface of the Earth. If it is thrown straight downward with an initial speed of 20 m/s, what will its kinetic energy be as it strikes the ground? (Ignore air resistance.) A. 400 J B. 200 J C. 800 J D. 600 J

D; A 2-kg object is at a height of 10 m above the surface of the Earth. If it is thrown straight downward with an initial speed of 20 m/s, Its kinetic energy will be 600 J as it strikes the ground. Use the conservation of energy equation:

A hammer is used to drive a nail into a board. Work is done in the act of driving the nail. Compared to the moment before the hammer strikes the nail, the mechanical energy of the hammer after its impact will be: A. less, because work has been done on the hammer. B. greater, because work has been done on the hammer. C. greater, because the hammer has done work. D. less, because the hammer has done work.

D; A hammer is used to drive a nail into a board. Work is done in the act of driving the nail. Compared to the moment before the hammer strikes the nail, the mechanical energy of the hammer after its impact will be less, because the hammer has done work. The hammer has done positive work on the nail, giving it enough energy to move into the wood. Thus, some of the hammer's original kinetic energy is lost, leaving it with less mechanical energy.

A simple pendulum consisting of a 1-kg bob connected to a rigid rod 5 m long is brought to an angle of 90° from vertical, and then released (as shown below). Assuming that the rod is massless and that the acceleration due to gravity is 10 m/s2, what will be the speed of the bob at its lowest point? A. 30 m/s B. 40 m/s C. 20 m/s D. 10 m/s

D; A simple pendulum consisting of a 1-kg bob connected to a rigid rod 5 m long is brought to an angle of 90° from vertical, and then released (as shown below). Assuming that the rod is massless and that the acceleration due to gravity is 10 m/s2, the speed of the bob at its lowest point is 10 m/s.

An erg is a unit of energy equal to 1 g·cm2/s2. The conversion between ergs and joules is therefore 1 joule = x ergs. What is x? A. 10^-7 B. 10^-5 C. 10^5 D. 10^7

D; An erg is a unit of energy equal to 1 g·cm2/s2. The conversion between ergs and joules is therefore 1 joule = x ergs, then x is 10^7. The joule uses kg for mass and m/s for speed, while the erg uses g (grams) for mass and cm/s for speed. Converting kg to g and m2 to cm2, we find:

An object is being pulled along the ground by a 50-N force directed 45° above the horizontal. Approximately how much work does the force do in pulling the object 8 m? A. 400 J B. 620 J C. 100 J D. 280 J

D; An object is being pulled along the ground by a 50-N force directed 45° above the horizontal. The force must do approximately 280 J to pull the object 8 m. The work done by a force F that acts over a displacement d is given by W = Fd cosθ, where θ is the angle between the vectors F and d. In this case, we compute that W = (50 N)(8 m)(0.7) = 280 J.

An object is released from rest at height h above the surface of the Earth, where h is much smaller than the radius of the Earth. Its speed is v m/s as it strikes the ground. At what height should this object be released from rest in order for its speed to be 2v when it strikes the ground? (Ignore air resistance; g denotes the magnitude of gravitational acceleration.) A. 4gh B. 2gh C. 2h D. 4h

D; An object is released from rest at height h above the surface of the Earth, where h is much smaller than the radius of the Earth. Its speed is v m/s as it strikes the ground. The object should be released from rest at 4h in order for its speed to be 2v when it strikes the ground. This is a proportion question (the MCAT loves proportions). Essentially it's asking "What should we do to the height if we want to double the impact speed?" All we need is to find the connection between height and impact speed. This is a "potential energy turns into kinetic energy" situation, so Therefore, if we want v to double, h must be increased by a factor of 2^2 = 4.

An object weighing 100 N is traveling vertically upward from the earth in the absence of air resistance at a constant velocity of 5 m/s. What is the power required to keep the object in motion? A. 20 W B. 0 W C. 200 W D. 500 W

D; An object weighing 100 N is traveling vertically upward from the earth in the absence of air resistance at a constant velocity of 5 m/s. 500 W is required to keep the object in motion. In order to maintain a constant velocity pointing upward, the net force in the vertical direction must be zero. Since the force of gravity is exerting a downward force of 100 N, a 100 N upward force must be provided to maintain no net vertical force. With this knowledge, we may use the equation P = Fv. This gives P = (100 N)(5 m/s) = 500 W.

An object's speed increases from 0 to 2 m/s, due to an amount of work W1, and then increases from 2 m/s to 4 m/s due to an amount of work W2. Which one of the following is true? A. Cannot be determined from the information given B. W1 = W2 C. W1 > W2 D. W1 < W2

D; An object's speed increases from 0 to 2 m/s, due to an amount of work W1, and then increases from 2 m/s to 4 m/s due to an amount of work W2, therefore W1 < W2 is the true statement. Use the work-energy theorem. The amount of work W1 equals the change in kinetic energy from speed 0 m/s to speed 2 m/s; therefore, W1 = Delta KE = (.5)m(4-0) = 2m W2 = Dela KE = (.5)m(16-4) = 6m Similarly, the amount of work W2 equals the change in kinetic energy from speed 2 m/s to speed 4 m/s: Therefore, W1 < W2.

If a projectile travels through air, it loses some of its kinetic energy due to air resistance. Some of this lost energy: A. causes the temperature of the air around the projectile to decrease. B. is found in increased potential energy of the projectile. C. decreases the temperature of the projectile. D. is found in increased kinetic energy of the air molecules.

D; If a projectile travels through air, it loses some of its kinetic energy due to air resistance. Some of this lost energy is found in increased kinetic energy of the air molecules. A loss of a projectile's kinetic energy can be attributed to either a gain in its gravitational potential energy, the transfer of heat, or both. In this case, we are explicitly told that the projectile loses some of its kinetic energy due to air resistance (friction). Thus, we conclude that some of the projectile's kinetic energy has been converted to heat. This causes, for one thing, the temperature of the air surrounding the projectile to increase; this means that the average kinetic energy of the air molecules increases.

A double tackle machine is a two-pulley system with the upper pulley attached to the ceiling and the lower pulley attached to a mass, with four strands of rope running between the two pulleys. What is the mechanical advantage of a double tackle machine lifting a mass M? A. 1 B. 2 C. 3 D. 4

D; The mechanical advantage of any simple machine such as a ramp, lever, or pulley system is defined as the ratio of the of the resistance force to the effort force. The resistance force is the weight Mg, whereas the tension in the four strands of rope is Mg / 4, so the mechanical advantage is Mg / (Mg / 4) = 4.

book sits at rest on a horizontal table. It is accelerated with a pushing force. Then, after it has moved a distance s, it is released. Soon after it comes to rest. From the beginning to the end of this problem, what is the net work done on the book? A. Fpush × s B. -Fpush × s C. -Ffriction × s D. 0 J

D; The net work done on the book as it moves across a horizontal surface (over which there is no change in gravitational potential energy) must equal the change in kinetic energy. Because the book started and ended at rest, ΔKE= 0 = Wnet.

An object slides down a 2-meter-long ramp that makes an angle of 60° with the horizontal. The mass of the object is 2 kg, and the coefficient of kinetic friction between the object and the ramp is 0.3. Calculate the work done by the normal force on the object. A. 20 J B. 6 J C. 3 J D. 0 J

D; The work done by the normal force N is automatically zero, since N is perpendicular to the displacement of the object. All the numbers given in the question are irrelevant.

Two objects, one with mass M1 and the other with mass M2 (where M2 > M1), rest at the top of an inclined plane. If the bottom of the incline is taken to be the zero of gravitational potential energy, then these objects have: A. different inertias but the same potential energy. B. the same inertia but different potential energies. C. the same inertia and the same potential energy. D. different inertias and different potential energies.

D; Two objects, one with mass M1 and the other with mass M2 (where M2 > M1), rest at the top of an inclined plane. If the bottom of the incline is taken to be the zero of gravitational potential energy, then these objects have different inertias and different potential energies. Since mass is the measure of inertia, the fact that the objects have different masses means they have different inertias. Now, potential energy is mass times g times height, so the fact that they're at the same height means they have different potential energies because their masses are different.

Total mechanical energy equation

E = U + K

How much power must be provided to a model rocket of mass 50kg to keep ir moving upward at a constant speed of 40 m/s?

F = mg F = (50kg)(10N/kg) F = 500N P = Fv P = (500N)(40m/s) P = 20,000 W or 20kW

Due to frictional effects within the car and air drag, a car's tires must exert a force of 10^4 N on the road to maintain a constant speed of 67 mph (30 m/s). What power is being transferred by the transmission to the tires at this speed? A. 0 W B. 3 × 10^4 W C. 3 × 10^5 W D. 6.7 × 10^5 W

For a constant force parallel to velocity, power is given by P = Fv. In this case that yields P = 10^4 N × 30 m/s = 3 × 10^5 W.

A force of magnitude 40 N pushes on an object of mass 8 kg through a displacement of 5 m for 10 seconds. What is the power provided by this force?

W = Fd cos theta W = (40N)(5m) W = 200 J P = 200J / 10s P = 20W

What are simple machines?

Inclined Plane, Wedge, Screw, Lever, Wheel and Axle, and Pulley

Impulse equation

J=Ft

Work units

Joules (Nm)

Kinetic energy equation

KE=1/2mv^2 m = mass v = velocity

Conservation of total mechanical energy equation (with outside forces equation

KEi + PEi + W = KEf + PEf

Conservation of total mechanical energy (no nonconservative forces) equation

KEi + PEi = KEf + PEf

Angular momentum equation

L = lmv = lw L = angular momentum l = lever arm (or radius) m = mass v = velocity l = rotational intertia w = angular velocity

Mechanical advantage equation

MA = F(resistance)/F(effort) = d(effort)/d(resistance)

When F is parallel to d, we can use the power equation

P = Fv

A car of mass 2000 kg accelerates from rest to a speed of 30 m/s in 9 seconds. Given that the engine does a total of 900,000J of work, what is the average power output of the car's engine?

P = W/t P = 900,000 J / 9s P = 100,000 W

Power equation

P=W/t p = power W = work t = time

Gravitational potential energy equation

PE = mgh

A 2-kg block slides down a 3-meter-long, frictionless 30° incline. If the block started from rest at the top of the incline, with what speed does it reach the bottom? A. 3.6 m/s B. 2.7 m/s C. 7.1 m/s D. 5.5 m/s

Since gravity is the only force doing work, we can use Conservation of Mechanical Energy: potential energy turns into kinetic energy. mgh = (1/2)mv2, so . The height of the plane is the length times sin 30° = 3 × (0.5) = 1.5 m. Therefore the speed at the bottom is which is approximately 5.5 m/s.

What is the effort force?

The force applied to a machine

An object of mass 10kg is moving at a speed of 9 m/s. How much work must be done on this object in order to stop it?

W = Kefir - KEi W = 0 - 1/2mv^2 W = -(.5)(10)(81) W = -405J

What are perfectly inelastic collisions?

The objects are stuck together afterwards

What is momentum?

The product of an object's mass and velocity

What is an elastic collision?

Total momentum and KE are conserved

What are inelastic collisions?

Total momentum is conserved but KE is not

Elastic potential energy equation

U = 1/2 k x^2 k = spring constant x = position

Efficiency equation

Useful energy output / total energy input

If the energy lost to friction when pulling this mass of 80kg up to a height of .5m is 100J, what was the efficiency of this system?

W = (80)(10)(.5) W = 400J E(input) = 400J + 100J = 500J E = 400J / 500J = 80%

A 5kg box starting at rest slides down a 3m high, 5m long ramp, reaching the ground with a seed of 4m/s. How much mechanical energy was lost due to friction?

W = -110J

A young child is sliding down a hill at an incline of 30 degrees on a sled with a total combined mass of 10kg. If the coefficient of friction between the hill and the sled is .3, and the length of the hill is 50m, how much work has been done by gravity when the child reaches the bottom of the hill? A. 1000 J B. 2500 J C. 3535 J D. 4330 J

W = F*d W = (m)(g)(d)(cos 60) W = (10)(10)(50)(1/2) W = 2500J

A ball of mass 2 kg is dropped from a height of 100m. As it falls, the ball feels an average force of air resistance of magnitude 4N. What is the ball's speed as it strikes the ground?

W = Fd W = 4N * 100m W = 400J KEi + PEi + W = KEf + PEf PEi + W = KEf mgh - W = 1/2mv^2 (2)(10)(100) - 400J = 1/2(2)v^2 2,000 J - 400J = v^2 1600 = v^2 v = 40 m/s

If the braking force is 20,000 N, how far will the car travel before stopping?

W = Fd d = W/F d = -2 x 10^5 J / -2 x 10^4 N d = 10m

How much work is required to stop a 1,000 kg car moving at 20 m/s?

W = delta KE W = KEf - KEi W = 0 - 1/2mv^2 W = -(.5)(1,000kg)(20m/s)^2 W = -2 x 10^5 J

Work done by gravity equation

W(g) = mgh

Work energy theorem equation

W=(delta)KE

Power units

Watts 1 W = J/s (kg*m^2/s^3)

What is the work-energy theorem?

When net work is done on/by system, the KE of the system will change by the same amount

What is work?

a measure of the how much force contributes to the displacement of an object

What is potential energy?

energy of position

What is impulse?

change in momentum

Is gravity a conservative or non-conservative force?

conservative

What is kinetic energy?

energy of motion

Ball 1 and ball 2 are rolling toward each other at the same speed, 5 m/s. Ball 1 has a mass of m1 = 8 kg, and ball 2 has a mass of m2 = 2 kg. After the collision, ball 1 is observed to move with a velocity of 2 m/s in the same direction as v1. What is the velocity of ball 2 after the collision?

m1v1 + m2v2 = m1v1' + m2v2' (8)(5) + (2)(-5) = (8)(2) + (2)(x) 40 - 10 = 16 + 2x 30 = 16 + 2x 14 = 2x x = 7m/s

Ball 1 rolls with an initial velocity v1 = 5 m/s towards ball 2, which is initially at rest. Ball 1 has a mass of 1kg, and ball 2 has a mass of 4 kg. After the collision, ball 2 is observed to move with a velocity of v2' = 2 m/s. What is the velocity of ball 1 after the collision?

m1v1 + m2v2 = m1v1' + m2v2' m1v1 + 0 = m1v1' + m2v2' (1)(5) = (1)(x) + (4)(2) 5 = x + 8 x = -3 m/s

Ball 1 and Ball 2 are rolling toward each other at the same speed, 5 m/s. Ball 1 has a mass of 8kg and ball 2 has a mass of 2 kg. After the collision, Ball 1 and Ball 2 stick together and slide frictionlessly across the table. What is their common velocity after the collision?

m1v1 + m2v2 = m1v1' + m2v2' m1v1 + m2v2 = (m1 + m2)v (5)(8) + (2)(-5) = (8 + 2)v 40 - 10 = 10v 30 = 10v v = 3 m/s

If cosine theta is between 90 and 180, work is

negative

Nonconservative forces typically do ______ work

negative

When work is done BY the system, energy is

negative

Is friction a conservative or nonconservative force?

nonconservative

A car whose mass if 2,000 kg is traveling at a velocity of 15 m/s due east. What is its momentum? How does its momentum compare to that of a car whose mass is 2,000 kg traveling at a velocity 15 m/s due west?

p = mv p = (2 x 10^3)(1.5 x 10^1) p = 3 x 10^4 kgm/s east p = -3 x 10^4 kgm/s west

A particle of mass 3kg moves with a speed of 5 m/s around a circle of radius 60cm. What is the magnitude of its momentum? Its angular momentum?

p = mv p = (3kg)(5m/s) p = 15 kgm/s L = rmv L = (.6)(3)(15) L = 9 kgm^2/s

momentum equation

p=mv

If theta is between 0 and 90, work is

positive

When work is done ON the system, energy is

positive

Is kinetic energy scalar or vector?

scalar

Is work scalar or vector?

scalar

When potential energy is less than 0, it is

spontaneous

What is resistance force?

the force applied by the machine

What is power?

the rate at which work is done

What is efficiency?

the ratio of output work to input work

A roller-coaster car drops from the rest down the track and enters a loop. If the radius of the loop is R, and the initial height of the car is 5R above the bottom of the loop, how fast is the car going at the top of the loop? Assume that R = 15m and ignore friction.

v = 30m/s

Is momentum a vector or scalar quantity?

vector

The area under a force vs displacement curve gives what?

work done by that force