Physics Chapter 7

How much energy is 1 kW-hr?

(1 kW-hr)(1000 W / kW)(3600 s / hr) = 3.6x106 J (Remember that a W is 1 J/s) This equation means that someone who is charged 0.22 $/kW-hr is really paying $0.22 for every 3.6 million Joules of energy.

A 50 kg box is initially at rest. A student applies a constant horizontal force of 40 N to the box for a distance of 4.5 m. What is the speed of the box at the end of this motion? (Assume no energy is lost due to friction.)

0 + W = K Fd = ½ mv^2 (40 N)(4.5 m) = ½ (50 kg)(v^2) v = 2.7 m/s

A 10 g bullet is fired horizontally out of a rifle with a 0.8 m long barrel. The speed of the bullet when it emerges from the barrel is 500 m/s. What was the average force of the exploding gases on the bullet while it traveled through the barrel?

0 + W = K Fd = ½ mv^2 F (0.8 m) = ½ (0.01 kg)(500 m/s)^2 F = 1563 N

What minimum power rating would a car engine need in order to go from 0 to 60 mph (0 to 26.8 m/s) in 6.0 seconds if the car has a mass of 1350 kg?

0 + W = K W = ½ mv^2 W = ½ (1350 kg)(26.8 m/s)^2 W = 484812 J P = W / t P = (484812 J) / (6.0 s) P = 80802 W Or (80802 W)(1 hp / 746 W) = 108 hp

A student connects a 500 kg container to a rope and pulley system and then connects the rope to a motor. If the student wants to lift the container 15.0 m in 2.0 min, what average power rating is required?

0 + W = Ug W = mgh W = (500 kg)(9.8 N/kg)(15.0 m) W = 73500 J P = W / t P = (73500 J) / ((2.0 min)(60 s / 1 min)) P= 613 W

1 Joule=

1 J= 1 N*m= 1 kg* m^2/s^2

1 Watt=

1 W= 1 J/s

1 horsepower=

1 hp= 746 Watts

Conservation of mechanical energy equation

1/2 mvi^2 + mghi + 1/2 kxi^2 = 1/2 mvf^2 + mghf + 1/2kxf^2, where "i" is initial and "f" is final.

1 calorie=

4.184 joules (1 kcal=4184 joules)

A 6.0 kg box moving with an initial speed of 15 m/s slows to a final speed of 8 m/s over a distance of 4.5 m due to friction. What is the friction force on the box?

Again, you can solve this using a combination of forces and kinematics or you can use energy. The box has kinetic energy at the start and less kinetic energy at the end. Where did this energy go? The box performed work on the surroundings through friction. energy start +/- work = energy end Kstart - W = Kend ½ mv2 - Fd = ½ mv2 ½ (6.0 kg)(15 m/s)2 - F (4.5 m) = ½ (6.0 kg)(8 m/s)2 F = 107 N

A 0.3 kg box is held against a spring causing the spring to compress 0.15 m. The spring has a spring constant of 400 N/m. When released, the spring will launch the box vertically. What is the speed of the box when it has reached a height of 0.5 m?

All of the energy is initially stored in the spring as spring potential energy. At the end, the energy has been transferred to the box. Some of this energy is gravitational potential energy and some of the energy is kinetic. energy start = energy end Us = Ug + K ½ kΔx2 = mgh + ½ mv2 ½ (400 N/m)(0.15 m)2 = (0.3 kg)(9.8 N/kg)(0.5 m) + ½ (0.3 kg)(v2) v = 4.5 m/s

A student attaches a 2.0 kg mass to a 1.2 m string and pulls the string back to an angle of 25° relative to the vertical. The student releases the mass, and it swings back and forth. What is the speed of the mass when it reaches the bottom of the swing?

All the energy at the beginning is gravitational potential energy. This is converted entirely to kinetic energy at the bottom of the swing. energy start = energy end Ug = K mgh = ½ mv2 You need to solve for the initial height of the mass relative to the bottom of the swing. h = (1.2 m) - (1.2 m) (cos 25) = 0.11 m Then back to the energy conservation equation: (2.0 kg)(9.8 N/kg)(0.11 m) = ½ (2.0 kg)(v2)v = 1.5 m/s

A 0.5 kg box is pressed against a spring causing the spring to compress 25 cm. When the spring is released, the box will be launched vertically into the air. What spring constant is required if the maximum height the box will reach is 1.5 m?

At the start of the process, all the energy is stored in the spring as spring potential energy. At the end of the process, all the energy has been transferred to the box, which is stored as gravitational potential energy. energy start = energy end Us = Ug ½ kΔx2 = mgh ½ (k)(0.25 m)2 = (0.5 kg)(9.8 N/kg)(1.5 m) k = 235 N/m

A 2.0 box is held against a spring causing the spring to compress 20 cm. How fast will the box move if the spring is released? The spring constant is 300 N/m.

At the start of this process, all of the energy is stored in the spring as spring potential energy. At the end of the process, all of the energy has been transferred to the box, which is stored as kinetic energy. energy start = energy end Us = K ½ kΔx2 = ½ mv2 ½ (300 N/m)(0.2 m)2 = ½ (2.0 kg)(v2) v = 2.4 m/s This assumes that no energy is transferred from the box to the ground through friction.

What is the cost to run a single 100 W light bulb for 30 days if the electric company charges $0.18 for every kW-hr?

Cost = (0.18 $/kW-hr)(1 kW / 1000 W)(100 W)(24 hr / 1 day)(30 days) Cost = $12.96

Change=

Delta (change)= final-initial

A spring with a spring constant of 200 N/m is stretched 25 cm. How much spring energy is stored in the spring? How much work was done on the spring?

Don't forget to convert cm to m. Us = ½ kΔx2 Us = ½ (200 N/m)(0.25 m)2 Us = 6.25 J

What is energy?

Energy is a way of predicting how much change a system can experience. The formal definition is that energy is the "ability to do work."

A student plans to use a motor to pull a 10.0 kg bucket of water out of a 100 m well. If the motor has a power rating of 2.0 hp, how long will it take to lift the water out of the well?

For this problem you'll consider 0 height to be the bottom of the well. This means the bucket starts with no energy and ends with gravitational potential energy. The motor performs work on the bucket. energy start +/- work = energy end 0 + W = Ug W = mgh W = (10.0 kg)(9.8 N/kg)(100 m) W = 9800 J P = W / t (2 hp)(746 W/hp) = (9800 J) / t t = 6.6 s

Hooke's Law (equation)

Fs=k[delta]x, where Fs is the force applied to the spring, k is the spring constant and delta x is the distance the spring was stretched or compressed

Defining work done on systems in terms of change in energy

If the system gains energy, then you say that work is done by the surroundings on the system. If a system loses energy, then you say that work is done by the system on the surroundings.

t takes a 1000 kg car 4.0 seconds to slow from a speed of 20 m/s to 15.0 m/s. What power is dissipated by the car?

K - W = K (1/2 mv^2)start - W = (1/2 mv^2)end ½ (1000 kg)(20 m/s)^2 - W = ½ (1000 kg)(15 m/s)^2 W = -87500 J P = W / t P = (87500 J) / (4.0 s) P = 2.2x10^4 W

A 0.5 kg box sliding along a frictionless surface slides up a frictionless hill. What is the maximum height of the box if the initial speed of the box is 10 m/s?

K = Ug ½ mv^2 = mgh ½ (0.5 kg)(10 m/s)^2 = (0.5 kg)(9.8 N/kg)(h) h = 5.1 m

A student throws a ball straight up into the air with a starting speed of 15 m/s. What is the speed of the ball after it has traveled upward 2.0 m? (Ignore air resistance.)

K = Ug + K (1/2 mv^2)start = mgh + (½ mv^2)end ½ (15 m/s)^2 = (9.8 N/kg)(2.0 m) + ½ (v^2) v = 13.6 m/s

How much kinetic energy does a 700 kg car have if it moves at a speed of 30 m/s?

K = ½ mv2 K = ½ (700 kg)(30 m/s)2 K = 315000 J or 315 kJ

What is the kinetic energy of a 1100 kg car traveling at a speed of 20 m/s?

K = ½ mv^2 K = ½ (1100 kg)(20 m/s)^2 K = 2.2x10^5 J

Conservation of mechanical energy defined

KE+PE=constant, or KEi+PEi=KEf+PEf for all conservative forces (such as spring force and gravitational force).

Kintetic energy (KE=)

KE= 1/2 mv^2, where KE is translational kinetic energy, m is mass and v is velocity.

List the three mechanical forms of energy and their associated equations.

Kinetic energy is the energy of motion: K = ½ mv^2. Gravitational potential energy is the energy due to an object's height in a gravity field: Ug = mgh. Spring potential energy is the energy stored in a spring when it is stretched or compressed: Us = ½ kΔx2

How much work does a force do on a system if it is perpendicular to the displacement of the system?

None. W=Fd cos(theta), so if theta is 90 degrees (perpendicular), cos theta is 0, and therefore work is 0.

A student uses a 1500 W space heater for 8.0 hours at night. How much energy does the heater use?

P = W / t (1500 W) = W / ((8.0 hours)(3600 s / 1 hr)) W = 4.32x10^7 J or 43.2 million Joules of energy

A typical hair dryer has a power rating of 1800 W. How much energy does it take for a student to dry her hair if she uses the dryer for 12 minutes?

P = W / t (1800 W) = W / ((12 min)(60 s / 1 min)) W = 1.3x106 J Be careful not to confuse the W for the units of power with the W for the symbol representing work.

What force does a car engine apply in order for a car to travel at a constant speed of 30 m/s if the motor is rated for 160 hp?

P = W / t P = Fd / t Since v = d / t (as long as v is constant) P = Fv (160 hp)(746 W / 1 hp) = F (30 m/s) F = 3979 N

Power (P=)

P=W/t, where P is power in units of watts, W is the work transferred in or out of the system, and t is the time it took to transfer the energy

What is the definition of power? What are the units of power?

Power is the rate at which energy is transferred into or out of a system. P = W / t The SI units are Joules per second or J/s which is the same as a watt (W). This can be converted to units of horsepower (hp).

Problem-solving strategy for problem dealing with energy

Step 1. Determine the system of interest and identify what information is given and what quantity is to be calculated. A sketch will help. Step 2. Examine all the forces involved and determine whether you know or are given the potential energy from the work done by the forces. Then use step 3 or step 4. Step 3. If you know the potential energies for the forces that enter into the problem, then forces are all conservative, and you can apply conservation of mechanical energy simply in terms of potential and kinetic energy (KEi + PEi = KEf + PEf) Step 4. If you know the potential energy for only some of the forces, possibly because some of them are nonconservative and do not have a potential energy, or if there are other energies that are not easily treated in terms of force and work, then the conservation of energy law in its most general form must be used (KEi + PEi + Wnc + OEi = KEf + PEf + OEf) . In most problems, one or more of the terms is zero, simplifying its solution. Do not calculate Wc , the work done by conservative forces; it is already incorporated in the PE terms. Step 5. You have already identified the types of work and energy involved (in step 2). Before solving for the unknown, eliminate terms wherever possible to simplify the algebra. For example, choose h = 0 at either the initial or final point, so that PEg is zero there. Then solve for the unknown in the customary manner. Step 6. Check the answer to see if it is reasonable. Once you have solved a problem, reexamine the forms of work and energy to see if you have set up the conservation of energy equation correctly. For example, work done against friction should be negative, potential energy at the bottom of a hill should be less than that at the top, and so on. Also check to see that the numerical value obtained is reasonable. For example, the final speed of a skateboarder who coasts down a 3-mhigh ramp could reasonably be 20 km/h, but not 80 km/h.

Which of the following students exerts more power? Student A carries a 40 kg box up a 4.0 m tall set of stairs in 10.0 seconds. Student B carries a 20 kg box up a 4.0 m tall set of stairs in 4.0 seconds. Explain.

Student B exerts more power. He or she carries half the mass up the same height in less than half the time. Quantitatively: P = W / t In both cases, the work is equal to the gravitational energy gained or mgh. Then: P = (mgh) / t Student A: P = (40 kg)(9.8 N/kg)(4.0 m) / (10 s) = 156.8 W Student B: P = (20 kg)(9.8 N/kg)(4.0 m) / (4.0 s) = 196 W

Explain why it is NOT appropriate to use the equation W = F//d when considering springs.

The above equation is only accurate if the force remains constant during a displacement. Different amounts of spring stretch require different forces. Since the force is not constant, the above equation is not accurate. If you want to consider a force on a spring, you must use Hook's Law: F = kΔx

A 1200 kg car travels from rest to a final speed of 30 m/s in 4.0 seconds. What is the power rating of the car's engine?

The car started with no energy and ended with kinetic energy. The motor performed work on the car. energy start +/- work = energy end 0 + W = K W = ½ mv2 W = ½ (1200 kg)(30 m/s)2 W = 540000 J P = W / t P = (540000 J) / (4.0 s) P = 135000 W Since it's unusual to talk about a car's motor in terms of watts: P = (135000 W)(1 hp / 746 W) = 181 hp

A 65 kg bungee jumper starts from rest at a height of 12.0 m. She jumps off the platform and reaches the lowest point of her motion (maximum bungee cord stretch) at 2.5 m above the ground. If the bungee cord has a spring constant of 250 N/m and an un-stretched length of 3.0 m, how much energy was lost to the surroundings during the initial fall?

The energy before the jump is all gravitational. The energy at the bottom of the jump is a combination of spring energy in the cord and gravitational potential energy for the jumper. You are solving for the work done by the jumper on the surroundings. The bungee cord experiences an amount of stretch equal to the difference between the total height and a combination of the unstretched length and final distance from the ground: Spring stretch = (12 m) - (3.0 m) - (2.5 m) Δx = 6.5 m Energy start +/- work = energy end Ug - W = Ug + Us mghstart - W = mghend + ½ kΔx2 (65 kg)(9.8 N/kg)(12 m) - W = (65 kg)(9.8 N/kg)(2.5 m) + ½ (250 N/m) (6.5 m)2 W = 770 J (The answer is a positive value because you already showed the work was negative by subtracting the term in the energy conservation equation.) This change in energy (work) could be caused by internal rope friction, air resistance, etc.

A student places a box on top of an inclined ramp that has a height of 0.8 m. The ramp's angle of incline is 20° and friction is negligible. What is the speed of the box when the box reaches the bottom of the ramp?

The energy starts in the box as gravitational potential energy. Even though the box is on a surface, it is still above zero height. This energy is converted to kinetic energy at the end. energy start = energy end Ug = K mgh = ½ mv2 (9.8 N/kg)(2.0 m) = ½ (v2) v = 6.3 m/s Notice how you did not need to use the given angle in the last example problem. The angle of the ramp is not important when considering gravitational potential energy since it is based on the object's height and not the object's overall displacement.

A student applies a horizontal force of 40 N to a box causing the box to move. The student starts to get tired after 10 m, and the applied force decreases to 20 N over the next 5 m. How much work was performed on the box?

The energy transferred to the box (work) is equal to the area under the above graph. You can break this area into a series of shapes and solve for those areas. W = area W = (40 N)(10 m) + (20 N)(5 m) + ½ (20 N)(5 m) W = 550 J

How do you solve for work?

The following are several ways to solve for work: Solve for the difference in a system's final and initial energy (i.e., ΔK or ΔU). Solve using the equation W = F//d (for constant forces only). Solve for the area under an F vs d graph (for any application of force).

Work-energy theorem (defined)

The work-energy theorem states that the net work done by all forces acting on a system equals its change in kinetic energy.

A particular computer has a power rating of 130 W. How much will it cost to run the computer for 24 hours if energy costs $0.26 for every kW-hr?

This is a unit conversion problem. You simply need to cancel out all of the units that are not related to cost. Cost = (0.26 $ / kW-hr)(1 kW / 1000 W)(130 W)(24 hours) Cost = $0.81

A 24 kg child descends a 5.0 m high slide and reaches the ground with a speed of 2.8 m/s. How much energy was dissipated due to friction in the process?

Ug - W = K mgh - W = ½ mv^2 (24 kg)(9.8 N/kg)(5.0 m) - W = ½ (24 kg)(2.8 m/s)^2 W = 1082 J

A student holds a 2.0 kg mass 3.0 m above a spring with a spring constant of 500 N/m. The student releases the mass from rest. What is the speed of the mass just before it hits the spring? How much will the spring compress when the mass hits it? How much would the spring compress if, instead of dropping the mass, the student had thrown the mass downward with a starting speed of 3.0 m/s?

Ug = K mgh = ½ mv^2 (2.0 kg)(9.8 N/kg)(3.0 m) = ½ (2.0 kg)(v^2) v = 7.7 m/s There is no need to include the negative since the question asked for speed and not velocity. Ug = Us mgh = ½ kΔ^2 (2.0 kg)(9.8 N/kg)(3.0 m) = ½ (500 N/m)(Δx^2) Δx = 0.48 m K + Ug = Us ½ mv^2 + mgh = ½ kΔ^2 ½ (2.0 kg)(3.0 m/s)^2 + (2.0 kg)(9.8 N/kg)(3.0 m) = ½ (500 N/m)(Δx^2) Δx = 0.52 m

A 500 kg rollercoaster car starts from rest at the top of a 15.0 m hill. The car travels down the hill, around a loop, and then travels up a smaller 5.0 m tall hill. What is the speed of the car when it reaches the top of the second hill? (Ignore friction and air resistance.) The second hill has a flat straight-away that ends with a very large spring. How much will the car compress the spring if the spring constant is 3000 N/m?

Ug = K + Ug (mgh)start = ½ mv^2 + (mgh)end (500 kg)(9.8 N/kg)(15.0 m) = ½ (500 kg)(v^2) + (500 kg)(9.8 N/kg)(5.0 m) v = 14 m/s Ug = Ug + Us (mgh)start = (mgh)end + ½ kΔx^2 (500 kg)(9.8 N/kg)(15.0 m) = (500 kg)(9.8 N/kg)(5.0 m) + ½ (3000 N/m)(Δx^2) Δx = 5.7 m That's a pretty large spring if its compression is nearly 6 m.

A student attaches a 1.5 kg mass to a spring. When the spring is released, the mass will fall. The student places a raw egg at a position 0.70 m below the starting point of the mass. What spring constant is needed so that the mass will fall, tap the egg without breaking it, and then bounce back up?

Ug = Us mgh = ½ kΔx^2 (1.5 kg)(9.8 N/kg)(0.7 m) = ½ (k)(0.7 m)^2 k = 42 N/m

What is the gravitational potential energy of a 30 kg box at rest on the top of a hill that is 15 m tall?

Ug = mgh Ug = (30 kg)(9.8 N/kg)(15 m) Ug = 4410 J

A 50 kg object is held at a height 10 m above the ground. How much gravitational potential energy does the object have?

Ug = mgh Ug = (50 kg)(9.8 N/kg)(10 m) Ug = 4900 J or 4.9 kJ

A 70 kg student stands on top of a 5.0 m platform diving board. How much gravitational potential energy does the student have? How much work did it take for the student to travel from the ground to the top of the platform diving board?

Ug = mgh Ug = (70 kg)(9.8 N/kg)(5.0 m) Ug = 3430 J W = (mgh)final - (mgh)start W = (3430 J) - 0 W = 3430 J

Gravitational potential energy (Ug=)

Ug=mgh, where Ug is the gravitational potential energy, m is mass, g is gravitational field strength and h is the change in height of the system from the point at which its potential energy is 0 (at rest).

A 25 g object is pressed against a spring causing the spring to compress 5 cm. The spring constant is 300 N/m. If the object moves vertically when released, what is the maximum height the object will reach? What maximum height will the box reach if 0.1 J of energy is lost due to air resistance? The spring is turned horizontally and the same object is used to compress the spring by the same amount. If friction is negligible, what is the speed of the object when the spring is released? How much force did it take to compress the spring?

Us = Ug ½ kΔx^2 = mgh ½ (300 N/m)(0.05 m)^2 = (0.025 kg)(9.8 N/kg)(h) h = 1.5 m Us - W = Ug ½ kΔx^2 - W = mgh ½ (300 N/m)(0.05 m)^2 - (0.1 J) = (0.025 kg)(9.8 N/kg)(h) h = 1.1 m Us = K ½ kΔx^2 = ½ mv^2 ½ (300 N/m)(0.05 m)^2 = ½ (0.025 kg)(v^2) v = 5.5 m/s F = kΔx F = (300 N/m)(0.05 m) F = 15 N

What is the spring potential energy of a spring that is stretched 15 cm if its spring constant is 350 N/m? How much force is required for this stretch?

Us = ½ kΔx^2 Us = ½ (350 N/m)(0.15 m)^2 Us = 3.9 J F = kΔx F = (350 N/m)(0.15 m) F = 52.5 N

Spring/elastic potential energy (Us=)

Us= 1/2kx^2, where Us is the potential energy of the spring, k is the spring's force constant, and x is the distance that the spring is compressed or stretched along its length.

initial energy=final energy

Us=K ½ kΔx2 = ½ mv2

A student applies a constant 200 N force horizontally to a box initially at rest for a distance of 3.0 m. How much work was done on the box?

W = F//d W = (200 N)(3.0 m) W = 600 J In this scenario, the student transferred 600 J of energy into the box.

A student applies a constant 200 N at an angle 30° below the horizontal to a 40.0 kg box for a distance of 3.0 m. How much work did the student do on the box?

W = F//d W = (200 N)(cos 30)(3.0 m) W = 520 J The student applied the same force as before over the same distance but did less work on the box since only a portion of the applied force contributes to the motion of the box. Note that the mass of the box is not important for this problem.

A student pulls a 10.0 kg wagon with a constant 140 N force at an angle 45° above the horizontal for a distance of 3.0 m. How much work did the student do on the wagon? What type of motion is the wagon experiencing (e.g., constant speed, acceleration, etc.)

W = F//d W = Fcosθd W = (140 N)(cos 45)(3.0 m) W = 297 J There is not enough information available to describe the motion. You would need to know how the horizontal component of the student's force on the wagon compares to the friction force on the wagon (or any other force that may or may not be acting on the wagon).

A student pushes a 40 kg lawnmower from rest across a rough surface with a force of 200 N at an angle of 60 degrees below the horizontal. If the lawnmower is moved 3 meters, how much work did the student do on the lawnmower? The coefficient of kinetic friction is 0.3. How much work was transferred from the mower to the ground through the friction force? What is the speed of the lawnmower at the end of the 3 m of motion?

W = F//d W = Fcosθd W = (200 N)(cos 60)(3.0 m) W = 300 J Solve for the normal force: ΣFy = 0 Fy + Fg + N = 0 (200 N)(sin 60) - (40 kg)(9.8 N/kg) + N = 0 N = 219 N Then: W = F//d = -fd W = -μNd W = -(0.3)(219 N)(3.0 m) W = -197 J 0 + W gained from student - W lost to ground = K (300 J) + (-197 J) = ½ (40 kg)(v2) v = 2.3 m/s

A student applies a constant 200 N of force horizontally to a box for a distance of 5.0 m. How much work does the student do on the box?

W = Fd W = (200 N)(5.0 m) W = 1000 J

How much work is required to make a 1400 kg car increase its speed from 10 m/s to 20 m/s? What average force is required if the car travels 15 m during this speed change?

W = ΔK W = (1/2 mv^2)final - (1/2 mv^2)start W = ½ (1400 kg)(20 m/s)^2 - ½ (1400 kg)(10 m/s)^2 W = 210000 J = 210 kJ W = Fd (2.1x10^5 J) = (F)(15 m) F = 14000 N

A spring is stretched from an initial stretch of 5 cm to a final stretch of 10 cm. The spring constant is 800 N/m. How much work was done on the spring? What is the final force on the spring when it is at its 10 cm stretch?

W = ΔUs W = (½ kΔx^2)final - (½ kΔx^2)start W = ½ (800 N/m)(0.10 m)^2 - ½ (800 N/m)(0.05 m)^2 W = 3.0 J F = kΔx F = (800 N/m)(0.1 m) F = 80 N

Work (W=)

W = ∣F∣ (cos θ) I d∣ , where W is work, d is displacement of the system, theta is the angle between the force vector F and the displacement vector d, and F is the magnitude of force on the system. (Also written as W = Fd cos θ)

Work when force is parallel (W=)

W=F//*d, where W is work done on the system, F// is the component of the force that is applied parallel to the motion, and d is the distance travelled. (W=Fd* cos[theta] where cos 0=1) *can only be used if force applied is CONSTANT*

Work-energy Theorem

Wnet = 1/2 mv^2 − 1/2 mv0^2, where Wnet is the net work done on the system, mv is mass*velocity, and mv0 is the initial mass*initial velocity

Net Work (Wnet=)

Wnet= Fnet*d, where Wnet is net work done on the system, Fnet is the net force applied to the system, and d is the distance

Explain why work is not considered a form of energy.

Work is the amount of energy transferred in or out of a system; it is not a way to store energy.

What is work?

Work is the change in energy experienced by a system. If the system gains energy, then work is done on the system. If the system loses energy, then work is done by the system.

A student applies a constant horizontal 50 N force to a 25 kg box that is initially at rest. The student moves the box a distance of 3.5 m. What is the speed of the box at the end of the motion? (Assume friction is negligible.)

You could solve this problem using a combination of Newton's second law and kinematics or you can use energy. At the start, the box has no energy as it's motionless at zero height. At the end, the box has kinetic energy. Where did this energy come from? The student applied a force to the box, which means that she performed work on the box. Energy start +/- work = energy end 0 + W = K Fd = ½ mv2 (50 N)(3.5 m) = ½ (25 kg)(v2) v = 3.7 m/s

A 5.0 kg box is held at a height of 2.0 m. The box is released. What is the velocity of the box just before it hits the ground?

You could solve this problem using traditional kinematic equations; however, try this solution from an energy perspective. At the start of this process, all of the energy in the system is stored as gravitational potential energy in the box. At the end of the process, all of the energy in the system is stored as kinetic energy. energy start = energy end Ug = K mgh = ½ mv2 (5.0 kg)(9.8 N/kg)(2.0 m) = ½ (5.0 kg)(v2) v = -6.3 m/s You chose the negative value after taking the square root since the direction of motion is down. Notice that the mass cancels out in this energy balance; the mass of a falling object does not affect the motion.

Which of the following is NOT a mechanical form of energy? a) Nuclear b) Kinetic c) Spring potential d) Gravitational potential

a) Nuclear

Which of the following are NOT units used to measure energy? a) Joules b) Newtons c) BTU d) Calories

b) Newtons

Which of the following is a correct interpretation of the law of conservation of energy? a) Energy cannot be transferred into a system. b) Energy cannot be transferred out of a system. c) Energy cannot be created or destroyed. d) None of the above are correct.

c) Energy cannot be created or destroyed.

A 700 kg car slows from an initial speed of 30 m/s to 20 m/s. What was the car's change in kinetic energy?

ΔK = Kfinal - KinitialΔK = ½ mv2 - ½ mv2ΔK = ½ (700 kg)(30 m/s)2 - ½ (700 kg)(20 m/s)2ΔK = 175000 J or 175 kJ

A student starting at a height of 6.0 m carries an 8.0 kg box down a flight of stairs to a final height of 2.0 m. What was the box's change in gravitational potential energy?

ΔUg = Ug final - Ug initial ΔUg = mgh - mgh ΔUg = (8.0 kg)(9.8 N/kg)(2.0 m) - (8.0 kg)(9.8 N/kg)(6.0 m) ΔUg = -314 J The negative sign indicates that the box lost gravitational potential energy. This should make sense since the box is now at a lower height. It has less potential for motional energy.