Physics II: Motion, Force, Energy and Equilibrium
In an open system are work/heat and mass exchanged with the system?
Yes, both work/heat and mass are exchanged in an "open" system.
How does an inclined plane reduce the amount of force needed to do work?
a ramp or inclined plan reduces the force needed by increasing the distance over which force is applied. To push a mass up a ramp only the force pushing the mass down must be overcome= mgsinθ if W is constant and force is the same then distance must increase to decrease force. ↑distance=↓force : inversely proportional W=Fd
Finish this concept: Energy of system + surroundings before =
energy of system + surroundings after
Define weight.
gravitational force x mass or mg
If a 1kg blocks were stacked one upon the other starting at the surface of the earth and continuing forever into space, the blocks near the bottom of the stack would have: A. less gravitational PE than blocks the middle or blocks near the top of the stack B. Less gravitational PE than blocks at the middle and the same gravitational energy as blocks near the top of the stack. C. The same gravitational PE as all other blocks. D. More gravitational PE than blocks at the middle or blocks near the top of the stack.
gravitational potential energy= Ug=mgh A. the blocks at the bottom have less height. They all have the same mass and the same gravity so they would have less PE than the blocks at the top which have more height.
Define mass
measure of the object's inertia.
Describe the force that points in a direction that is parallel to the inclined plane.
mgsinθ The net force due to gravity (weight) and the normal force on an inclined plane= mgsinθ mg + Fn (mgcosθ) mgSinθ: slides down the incline
Describe these scenarios in a velocity vs. time graph: Slope Upward slope Downward slope Straight line Curved line Area beneath the curve
slope= instantaneous acceleration Upward slope=positive acceleration Downward slope= negative acceleration Straight line= constant slope= constant acceleration Curved line= ∆slope=∆acceleration Area under the curve=acceleration
What situation has to occur to keep torque constant while increasing the force?
t=Fl t=Fr t=Frsinθ To keep torque constant while increasing force the lever arm must decrease. Force and lever arm are inversely proportional if torque is constant. If one increases the other must decrease and vice/versa. Like with PV=nRT. If V↑ then P↓. If P↑ then V↓.
What is the magnitude or definition of the torque vector?
t=Frsinθ t=Fl F= force vector r= position vector θ= angle btw force and position vectors
define inertia
tendency for an object to remain in their present state of motion.
What is newton's first law of motion?
the law of inertia: an object in a state of rest or in a state of motion will tend to remain in that state unless acted on by an outside force.
Finish this concept: Total energy of a system=
the sum of all forms of energy in that system
4 kinematic formulas
x-x₀=v₀t + (1/2)at² vf-v₀=at vf²=v₀² + 2a(xf-x₀) v(avg)= (1/2)(vf+v₀)
What are the 4 units to look for in a kinematics equation?
x= distance/displacement v=velocity a=acceleration t=time ∆=change
Equation for velocity
∆displacement/time
Equation for speed
∆distance/ time
A hiker throws a rock horizontally off a cliff that is 40 m above the water below. If the speed of the rock is 30 m/s, how long does it take for the rock to hit the water? A. 3s B. 4s C. 5s D. 6s
(30 m/s)/ (10 m/s²) A. 3 seconds The height does not matter. This is a projectile so the velocity and the gravity that enacts on the velocity matter.
Finish the three rules of keeping track of energy: 1. Energy of a system + surroundings before = 2. Energy leaving a system= Energy entering a system 3. Total energy of a system=
=Energy of system + surroundings after =Energy entering the surroundings =Energy leaving the surroundings =the sum of all forms of energy in that system
How does a lever reduce the amount of force needed to do work?
A lever reduces the force needed to do a given amt of work by increasing the distance over which the force is applied. The lever is based on the principle of torque. t=Force x lever arm : inversely proportional if torque is constant then ↑lever arm=↓force.
If an object is motionless what forces are acting on the object and what type of equilibrium is created?
A motionless object has Normal force and Fgravity. The type of equilibrium is static equilibrium with a constant zero velocity -(not moving)
How does a pulley reduce the amount of force needed to perform work?
A pulley uses rope to increase the distance over which force acts. Force and distance are always inversely proportional. As distance increases force decreases. Tension is in dynamic equilibrium. The magnitudes of the upward and downward forces are equal. F=T 2T (2xropes)=mg T=mg/2
What two components does a velocity vs. time graph tell?
A velocity vs. time graph not only explains the velocity but also the acceleration. If questioned on what a velocity vs. time graph is showing first think about acceleration. If there is an upward slope then there is positive acceleration.
A particle moving forward in a straight line slows down at a constant rate from 50m/s to 25m/s in 2 sec. What is the acceleration of the particle? A. -12.5 B. -25 C. +12.5 D. +25
A. -12.5 m/s² 50-25=25 m/s (-25 m/s)/2 -12.5 m/s² The ∆ from 50 to 25 means deceleration and a ∆ in velocity. The 50 is positive and the 25 is negative since that is the slow down so the average velocity is -25 m/s. Divided by the time gives the rate of change in the velocity or acceleration
A skydiver jumping from a plane will accelerate up to a max velocity and no greater. This constant velocity is known as terminal velocity. Upon reaching terminal velocity, the net force on the skydiver is: A. 0 and skydiver is in equilibrium B. 0 and the skydiver is not in equilibrium C. equal to the weight of the skydiver and the skydiver is in equilibrium. D. equal to the weight of the skydiver and the skydiver is not in equilibrium.
A. 0 and the skydiver is in equilibrium. Constant velocity requires the upward forces=downward forces so weight=air resistance. This would mean that the net forces are 0 and by definition the skydiver is in equilibrium.
What is required for any change in motion- direction or speed? A. Acceleration B. Normal force C. displacement D. inertia
A. acceleration is required for any change in motion. a= (∆velocity)/(∆time) v=(∆displacement)/(∆time)
The rate of change in velocity is called:
Acceleration. acceleration= ∆velocity/time
A way to think: when thinking about net force, what other motion vector should I think about?
Acceleration. It is net force that creates acceleration and causes changes in motion. F=ma
For a system in equilibrium what has to be the sum of all the forces and torques acting on the system?
All the forces and torques have to equal each other, cancel out and leave no net force.
Newton's first law (law of inertia):
An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
A small metal ball (m = 0.5 kg) is launched off a spring at a 30° angle to the ground, as pictured in the diagram below. Friction and air resistance are negligible in this case. If the ball has a horizontal velocity of 17.3 m/s after one second of flight, what was its initial kinetic energy? Note that sin(30°) = 0.5 and cos(30°) = 0.866. A. 75 J B. 100 J C. 200 J D. 300 J
B. 100J Important info: Horizontal velocity remains constant thru-out projectile arcs as long as air resistance is ignored. so Vx=17.3 m/s all thru the arc. vx=vcos30=20m/s KE=½(0.5)(20)=100J
A car accelerates at a constant rate from 0 to 25 m/s over a distance of 25 m. Approx., how long does it take the car to reach the velocity of 25 m/s? A. 1s B. 2s C. 4s D. 8s
B. 2 s ∆v=v₀=0 vf= 25 m/s x= 25m ∆a=constant t=? x=(½)(vi + vf)t 25m=((0+25)/2)t t=2s
Which of the following describes a situation requiring no net force? A. a car starts from rest and reaches a speed of 80 km/hr after 15 seconds. B. A bucket is lowered from a rooftop at a constant speed of 2 m/s. C. A skater glides along the ice, gradually slowing from 10 m/s to 5 m/s. D. The pendulum of a clock moves back and forth at a constant frequency of 0.5 cycles/second.
B. A bucket is lowered from a rooftop at a constant speed of 2 m/s. Net force= acceleration so no net force= no acceleration A= rest to movement means acceleration was changing C= slowing from 10 to 5 m/s means deceleration so acceleration was changing. D. Pendulum movement changes= changes in motion= changes in velocity=changes in acceleration. B= constant speed= no acceleration= no net force.
A ball attached to a string on a tabletop is swung in a circle with a velocity of v. The string is threaded through a small hole in the table such that it can be pulled on to be shortened. If the string is loosened and allowed to extend to three times its original length, how will the ball's velocity change? A. The velocity of the ball will remain constant. B. The velocity of the ball will decrease to one-third of its previous value. C. The velocity of the ball will increase by a factor of 1.7. D. The velocity of the ball will increase by a factor of 3.
B. The velocity of the ball will decrease to 1/3 of its previous value. Info needed: Angular momentum which is momentum in a circular, spinning object like ball on the string. Angular momentum= L=mvR. L= angular momentum m=mass, v=velocity R=string length If L stays constant and m is constant then v and R. For L to be constant then v and R are indirectly proportional. So for one to decrease the other must increase. The string increased so the velocity must decrease.
A 90-g cell phone is unfortunately dropped directly off the roof of a one-story building. If the phone takes 1 s to hit the ground, what was its gravitational potential energy while still on the roof? A. 0 J B. 4.5 J C. 9 J D. 4500 J
B=4.5J Energy problem! ∆x=vit +½at² v=0 t=1 a=10 m/s² ∆x=5m PE=mgh= (0.09kg)(10m/s²)(5m)=4.5J
Jessie's high school physics class is running a "potato cannon" competition. The goal is simple: shoot a potato the greatest possible horizontal distance. Right now, Jessie's cannon shoots potatoes at a 30° angle from the ground with a total velocity of 14 m/s. What change(s) can Jessie make to increase the distance her potatoes travel? I. Increasing the total velocity to 18 m/s while keeping all other factors constant II. Decreasing the masses of the potatoes to make them fall more slowly III. Changing the angle to 45° with respect to the ground IV. Changing the angle to 90° with respect to the ground Note that cos 30 = 0.87, sin 30 = 0.5, cos 45 = 0.71, and sin 45 = 0.71. A. I only B. I and II only C. I and III only D. I, III, and IV only
C. Think of horizontal and vertical components. The total velocity world be horizontal and since dx=v₀xt increasing velocity increases distance (horizontal component) Changing the angle would affect the initial vertical velocity so making the angle larger would increase the initial velocity. Mass does not impact the time taken for a projectile to fall a certain distance.
A 12-kg bag of clothes is lifted until it has 360 J of potential energy. Approximately how long will it take to hit the ground if dropped? A. 0.3 s B. 0.6 s C. 0.8 s D. 6 s
C. 0.8s Key words: lifted, potential energy, dropped First find the height the bag was dropped at: 360J=(12kg)(10m/s²)(h) h=3m Since only vertical components are used then kinematics is needed. ∆x=v₀t + ½at² 3m=0 + ½(10)t² t²=0.6 The square root of a decimal is a slightly larger decimal so 0.8 fits.
A weather balloon travels upward for 6km while the wind blows it 10km north and 8km east. Approx., what is its final displacement from its initial position? A. 7km B. 10km C. 14km D. 20km
C. 14km This is a Pythagorean theorem question. 6² + 10² + 8²= 196. √(196)=14 Also, if adding the vectors 10 and 8 the ∑vector would be 2≤x≥18 and 14 fits this rule
A driver moving at a constant speed of 20 m/s sees an accident and hits the brakes. If the car decelerates at a constant rate of -5 m/s², how far does the car go before it comes to a stop? A. 10m B. 20m C. 40m D. 100m
C. 40m vf²=v₀² + 2a(∆x) 20²=0 + 2(-5m/s²)(∆x) 400=(-10m/s²)(∆x) 400/10=40m
A box starts from rest and slides 40m down a frictionless incline plane. The total vertical displacement of the box is 20m. How long does it take for the block to reach the end of the plane? A. 1s B. 2s C. 4s D. 8s
C. 4s vertical displacement= ∆x Missing variables: a and t mgsinθ = ma a=gsinθ sinθ= 20/40=½ a=(½)(10m/s²)= a=5m/s² x=v₀t +½at² 40m=0 + ½(5 m/s²)t² t=4s
A metal box with a glass bottom slides down a steep incline. The incline can be assumed to be frictionless, and no air resistance acts on the box as it slides. However, a substantial amount of gravitational potential energy is converted to kinetic energy. Which of the following statements about this process is true? A. Mechanical energy is conserved, but total energy is not. B. Neither total nor mechanical energy is conserved. C. Total energy and mechanical energy are both conserved. D. Total energy is conserved, but mechanical energy is not.
C. Mechanical energy= KE + PE Total energy= KE + PE (U=p+w) so mechanical and total energy or always conserved and in some cases interchangeable.
An airliner flies from Chicago to NY. Due to the shape of the earth, the airliner must follow a curved trajectory. How does the curved trajectory of the airliner affect its final displacement for this trip? A. The displacement is less that it would be if the airliner flew in a straight line to NY. B. The displacement is greater than it would be if the airliner flew in a straight line to NY. C The displacement is the same as it would be if the airliner flew in a straight line to NY. D. The final displacement of the airliner is zero.
C. The displacement is the same as it would be if the airliner flew in a straight line to NY. Displacement is path independent. It cares about the initial point and final point not the the path taken in btw.
If a student working with this mechanical system needed to halve the ball's velocity without lengthening the string, she should: A. introduce static friction between the ball and the tabletop. B. use a heavier string between the ball and the point of rotation. C. use a heavier ball. D. none of the above; manipulating the string's length is the only mechanism by which this goal can be achieved.
C. Use a heavier ball Angular momentum: L=mvR m:mass, v:velocity, R:string length. If L is constant and R is constant then only mass and velocity change; they are indirectly proportional so as one increase the other decreases. To half velocity (decrease it) then mass must increase.
All of the following describe the magnitude and direction of a vector EXCEPT: A. 10 m/s B. 10 m/s in a circle C. 20 m to the left D. 20 m straight up
C. Vectors follow a straight path.
An engine is supplied with 1 kg of a hydrocarbon fuel with a heat of combustion of 50 MJ/kg. The engine can consume exactly 1 kg of fuel per minute. If this engine can produce 150 kW of power for one minute, the efficiency of the engine is: A. 0.3%. B. 2%. C. 20%. D. 100%
C.20% Info: Efficiency=(energy out)/(energy in) Power= W/t= 150kW (150 x10³)= 1.5x10⁵ W (1.5x10⁵)/60s=9x10⁶J=9MJ energy out=9MJ energy in=50MJ 9/50 ≈ 10/50 ≈ 1/5 ≈ 0.2 = 20%
A spring powered dart-gun fires a dart 1m vertically into the air. In order for the dart to go 4m, the spring would have to be depressed: A. 2 x the distance B. 3 x the distance C. 4 x the distance D. 8 x the distance
Concept: spring potential energy conversion into gravitational potential energy ½k∆x²=mgh Initial energy=½k∆x² Final energy= mgh The dart begins at spring/elastic potential energy but is converted into gravitational potential energy. ½kx²=mg4 x=2
A 10 kg mass is in free fall with no air resistance. In order to slow the mass at a rate equal to the magnitude of g, an upward force must be applied with magnitude: A. 0 N B. 10 N C. 100 N D. 200 N
D. 200N Two forces are acting on the object: Weight and normal force. To slow the object the normal force must overcome the weight of the object so: weight= (10kg)(10m/s²)= 100N To slow object down normal force should be x≥ 100N which is D. 200N
Several manic gorillas are throwing objects from a skyscraper. At the exact same instant, one of the gorillas drops a 15-kg table directly downward, another throws a 20-kg refrigerator horizontally with a velocity of 10 m/s, and a third drops a 1-kg lunchbox straight down. If air resistance can be neglected, which object will hit the ground last? A. The table, due to its larger mass and negligible initial velocity B. The refrigerator, due to its substantial horizontal velocity C. The lunchbox, due to its smaller mass D. All three objects will reach the ground at the same time.
D. All three objects will reach the ground at the same time. Time in flight does not depend on mass. If two objects are dropped simultaneously from the same height they will hit the ground at the same time. Air resistance is the important component for falling objects but since it is negligible it doesn't matter.
There are 3 forces acting on an object. Two of the forces are of equal magnitude. One of these forces pulls the object to the north and one pulls to the east. If the object undergoes no acceleration, then in which direction must the third force be pulling? A. Northeast B. Northwest C. Southeast D. Southwest
D. Southwest. The center is the connection btw north and east and southwest would bring the system to have no acceleration.
The earth moves around the sun at approximately 30 m/s. Is the earth acceleration? A. no, because the acceleration is a vector. B. No, b/c the speed is constant C. Yes, b/c the speed is not constant D. Yes, b/c the velocity is not constant.
D. Yes, the velocity is not constant. The direction of the earth is constantly changing even though it is going in a circle.
A slingshot propels a rubber ball horizontally from a 30 m platform at 15 m/s. Approximately how far from the base of the platform will the ball land? A. 14 m B. 21 m C. 30 m D. 38 m
D= 38m Known: distance= 30m high Vo= 15m/s Projectile= horizontal and vertical components. Vertical: 30m (height), 10m/s² (gravity) ∆x=v₀t+(1/2)at² 30=0+(1/2)(10 m/s²)t² 30=5t² t²=6 t≈2.5 Horizontal components:15 m/s, time=2.5s, distance= horizontal in this case distance=v₀xt= (15m/s)(2.5s)= 37.5
What's the difference btw displacement and distance?
Displacement cares about the final position -> shortest path. Distance cares about the path taken and the amt of meters moved.
A large rock is tied to a rubber band and dropped straight down. As the rock falls, the rubber band gradually stretches, eventually bring the rock to a stop. Which of the following energy transfers is taking placed in this process? A. KE to PE to elastic PE B. KE to elastic PE to PE C. PE to elastic PE to KE D. PE to KE to elastic PE
Energy changing and transfers stationary the rock= PE once falling rock goes from PE to KE When rock stops the rubber band is stretched and has stopped so KE is converted into elastic PE.
Finish this concept: Energy leaving a system= Energy entering a system=
Energy leaving a system= energy entering the surroundings Energy entering a system=energy leaving the surroundings
Consider a positively-charged particle that is experiencing a force due to an external electric field. Which of the following are conserved for the particle? I. Potential energy II. Kinetic energy III. Total energy IV. Momentum
Energy/Matter is not created nor destroyed, it can only be transferred into something else. PE can change into KE and vise versa. Momentum can change (momentum= mass x velocity) III. Total energy is always conserved
A system that experiences no translational/rotational acceleration is said to be in:
Equilibrium A system in equilibrium experiences no net force, no net torque, and no translational/rotational acceleration.
Explain the law of universal gravitation.
F=G (m₁m₂/r²) As the radius btw the two masses increases the gravitational force between the two masses decreases. They each fell less of the other's force. As they come closer they each feel the other's force.
Newton's second law:
F=ma The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
A crate is to be lifted to a height of 3 m with the assistance of an inclined plane. If the inclined plane is non-ideal machine, which of the following statements is most likely true? A. the non-ideal inclined plane increases the force required and decreases the work that has to be done. B. the non-ideal inclined plane decreases the force required and increases the work. C. the non-ideal inclined plane increases the force and the work required. D. the non-ideal inclined plane decreases the force and the work required.
FACT: Machines reduce amt of force necessary for work. Non-ideal machines are still machines so they still make work easier but they increase the force.
A frictionless pulley system below reduces the force necessary to lift any mass by a factor of 3. How much power is required to lift a 30 kg object 2m in 60 seconds using this pulley system? A. 4 W B. 10 W C. 24 W D. 120 W
Factor of 3 is useless information. Power=? Work=? W=mgh Pulley systems use PE W=(30kg)(10m/s²)(2m)= 600N P=600 N/60s= 10 W When asked for height think about PE and in pulley systems mgh is a component. Work =energy transfer
T/F. An object moving at 10 m/s north one moment and 10 m/s east the next moment has not accelerated.
False. Any change in direction means something has changed in acceleration despite the same velocity. The object normally has to slow down and speed up again to turn direction.
T/F. Machines change the amount of work but not the amount of force.
False. Machines do NOT change work. Machines change force. The purpose of a machine is to reduce the force required to do the work.
An inventor designs a machine that he claims will lift a 30kg object with the application of only a 25N force. If the inventor is correct, what is the shortest possible distance through which the force must be applied for each meter that the object is raised? A. 5m B. 8m C. 12m D. 15m
Fd=mgh d=mgh/F d=(30kg)(10m/s²)(1)/(25N) d= 300/25N d=60/5 d=12 m Work is also energy transfer so this machine is transferring energy. Since something is being raised potential energy is the type being transferred but work is still being done so to find the distance: Work=PE
What is newton's third law?
For every action there exists an equal and opposite reaction.
Newton's third law:
For every action, there is an equal and opposite reaction.
Describe the normal force.
Force pushing against the object on the surface of another object (i.e inclined plane, earth). It is always perpendicular to the surface.
A parachute jumper reaches constant velocity. What forces are acting on the jumper and what type of equilibrium is created?
In the air there if Fgravity and Fair resistance. Once these forces are equal then velocity is constant since the jumper is still falling just the vertical displacement/time is constant. This is then dynamic equilibrium where there is constant, non-zero velocity.
Is torque a vector or scalar?
It is a vector and it goes clockwise or counter-clockwise.
Simply describe potential energy and kinetic energy along with their equations.
KE=½mv² PE=mgh KE is energy of motion PE is energy of position
A meteor with a mass of 1kg moving at 20 km/s collides with Jupiter's atmosphere. The meteor penetrates 100 km into the atmosphere and disintegrates. What is the average force on the meteor once it enter Jupiter's atmosphere? (ignore gravity) A. 2x10³ N B. 4x10³ N C. 8x10³ N D. 2x10⁵ N
Key words: disintigrates= vi=0 Ignore gravity= no PE so work is only KE W=Fd=F(100km)=½mf²-½mvi² ½mvi²=(½)(1kg)(20 km/s)²-(½)(1kg)(0) F(100x10³m)=(½)(1kg)(20x10³m) F=(½(400x10⁶))/(100x10³m)= 2x10³N
A 100 N force is applied at at 30° downward angle towards a 10kg box for 2 seconds. If the object is initially at rest, what is its final velocity? (ignore friction: sin30=0.5; cos30=0.87) A. 8.7m/s B. 1m/s C. 17.4 m/s D. 34.8 m/s
Key words: force applied, final velocity= w=fdcosθ and kinematics Draw the figure. w=fdcosθ → the box doesn't move so no d w=fcosθ= 100Ncos30=87N F=87N ma=87N=(10kg)(a)=87N=a=8.7 m/s² vf=v₀ + (8.7 m/s²)(2) vf=17.4
A wheelchair access ramp is to be designed so that 1000 N can be lifted to a height of 1 meter through the application of 50 N of force. The length of the ramp must be at least: A. 5m B. 10m C. 20m D. 100m
Key words: force, height, length Work in=Work out Fd=Fd (50N)(d)=(1000N)(1m) d=20m Machines use force to derive a certain output. So the machine is applying a certain amount of force (50N) so that a force of 1000N can be lifted 1m. So a force is being applied for a n outcome.
What minimum coefficient of kinetic friction must a flat, rough surface 10 m in length have if it is to stop a sliding box with a momentum of 140 kg m/s? A. 0.13 B. 0.25 C. 0.5 D. Not enough information is given.
Kinetic friction: energy needed to stop a moving object. D is correct. We need the velocity of the object to solve this problem. From the information mentioned, there is no way to discern whether this is a heavy object with a low velocity (for which we would need a small µ) or a light object with a high velocity (where a larger µ would be required).
A 24-kg lawnmower is sitting in a shed. It is then dragged along the ground with a constant force of 120 N. Assuming that no friction is present, how far has it traveled when its velocity reaches 30 m/s? A. 3 m B. 6 m C. 45 m D. 90 m
Known: m=24kg F=120N v=30 m/s What is the distance travelled. F=ma vf²=vi² +2a∆x 120N=(24kg)xa a=5m/s² 30²=0 + 2(5)∆x 900=10x 90=x D.
A small metal racecar is positioned at the top of a curved ramp and released. The mass of the car (m) is 2 kg, the length of the track (L) is 15 m, and the track's height (h) is 3 m, while µk = 0.1. (Note: assume the average frictional force during the drop is µmg.) With what velocity will the racecar leave the track? A. 5.5 m/s B. 6.5 m/s C. 15 m/s D. 30 m/s
Known: mass= 2kg L=15m h=3m uk=0.1 velocity once the car leaves track Energy transfer problem! At the end of the track the car has KE converted from PE but also converted into friction so KE=mgh-ukmgL ½2v²=(2)(10)(3)-(0.1)(2)(10)(15) v²=30 v≈5.5
Vector
Magnitude + direction i.e: displacement, velocity, acceleration
Scalar
Magnitude but no direction. i.e: distance and speed
A 10-kg sled rests on a 30° ramp with a coefficient of static friction of 0.5. An upward force is applied to the sled, parallel to the slope of the ramp, in incremental values until the sled begins to accelerate up the ramp. Approximately what minimum force is required to perform this action? A. 44 N B. 50 N C. 95 N D. 120 N
Minimum force = Fnet, it is the force required to overcome all the other forces keeping the sled stationary. In this case: Fapplied=Fg+Ffriction. The force applied is the force to overcome the net forces of gravity and static friction. Ffriction= uFgcos30 Force parallel to incline plane:Fgsin30 mgsin30°=(10kg)(10m/s²)(0.5)= 50N Fs=(umgcos30°)=(0.5)(10kg)(10m/s²)(0.8)=43.3N Fapplied= 50N + 43.3N =93.3N or C. 95N
Do velocity and acceleration vectors have to be in the same direction?
No. Something can be moving to the left but accelerating to the right this is because acceleration is what causes something to speed up or slow down since it is the rate of change of velocity.
Are work/heat and mass exchanged with the surroundings in an isolated system?
No. work and heat and mass are not exchanged with the surroundings in an "isolated" system.
What unit measures the rate of energy transfer and thus the amount of work done?
P=W/t P=∆E/t t= time during energy transfer ∆E= energy change of the system = q + w
In a displacement vs. time graph describe what each scenario represents: Slope Upward slope Downward slope Straight horizontal line Curved line Area underneath the graph
Slope= instantanous velocity or velocity at that pt in time. Upward slope= pos velocity Downward slope= neg velocity straight horizontal line= slope of zero Curved line= ∆slope, indicating a ∆velocity and ∆acceleration.
What does torque measure?
The ability of a force to produce rotational motion.
What is newton's second law of motion?
The change in that object's state of motion will be inversely proportional to the mass(m) of the object and directly proportional to the net force (F) acting on the object. F=ma
What 2 variables can increase torque?
The component of the force acting perpendicular to the position vector. The distance btw the point of application of the force and the point of rotation.
A 10-kg crate is set on a cargo ramp with a 30° incline. If the crate is to remain in place, which of the following statements must be true? A. The ramp's coefficient of kinetic friction must be at least 0.6. B. The ramp's coefficient of static friction must be at least 0.6 C. The coefficient of kinetic friction cannot be less than one, so the ramp must be adjusted to a smaller angle in order for the crate to remain at equilibrium. D. The coefficient of static friction cannot exceed one, so the ramp must be adjusted to a smaller angle in order for the crate to remain at equilibrium.
The question is asking about friction. Static friction= force that keeps the box in place. Kinetic friction= force that would slow down the box to a halt if it was moving. To keep the box in place static friction in at play. To keep the box in place Fs≥mgsinθ mgsin=force trying to push box down the slide that friction has to overcome mgsin30= (100)(0.5)=50N Fg=Usmgsin30 100=Us(50) 0.5=Us The static friction coefficient should be at least 0.6
A marathon runner is jogging in a loop around the city with a velocity of 2.5 m/s. If the loop is perfectly circular and has a diameter of 3 km, after how many seconds will the runner reach his starting position? A. 67s B. 2000s C. 4000s D. 40000s
The question is asking how long does it take for the runner to finish the track. The loop is perfectly circular so diameter=length. v=d/t distance is 3km →convert. 1000m/1km=3000m 2.5=(3000m)/t 3000/(2.5)=4000s
An automobile that was moving forward on a highway pulled over onto the exit ramp and slowed to a stop. While the automobile was slowing down, which of the following could be true? A. the velocity was positive and the acceleration was positive B. the velocity was negative and the acceleration was negative C. the velocity was positive and the acceleration was negative D. the velocity and acceleration had the same sign, either positive or negative.
The velocity vector was opposite the acceleration vector. Acceleration is ∆v/∆t so it causes the changes in velocity. So if something needs to slow down it decelerates which causes changes in the velocity. So the velocity was positive since the car was still moving and the acceleration was negative since the car was slowing down to a stop.
A rescue helicopter lifts a 50 kg rock climber by a rope from a cliff face. The rock climber is accelerated vertically at 5 m/s². What is the tension in the rope? A. 350N B. 500N C. 750N D. 1500N
The weight of the climber is 50kg x 10m/s²= 500N for the helicopter to lift the rock climber the upward force has to be greater than 750N. The tension in the rope is still F=ma so 50kg x 5m/s²= 250N so T=250N. The tension in the rope is the combination of the weight on the climber and the T of the rope=750N.
Describe Hooke's law (spring constant).
There is a deforming force that is created when something is bent out of shape. Mainly used for springs. F=-k∆x F=force k=unique constant to the spring ∆x= change in position
T/F. Distance is always positive.
True. Distance is a scalar quantity=magnitude but no direction. Positive/Negative= direction Direction would be always increasing no matter how long it goes unlike displacement which shows direction and magnitude so can be negative/positive.
Energy consumption in the home is generally measured in units of kilowatt hours. A kilowatt hour is equal to: A. 3600J B. 6000J C. 3,600,00J D. 6,000,000J
Unit conversion: kw=Power=energy transfer/time 1N=1J/s 1kW=10³ J/s 1kW=(10³ J/s)(60 s/min)(60 min/hr) =3600x10³J=3,600,000J
When moving a box across the floor what is the vertical component and the horizontal component?
Vertical component is 0 since there is 0 vertical displacement or the box doesn't move up. The box is moving across the floor so there is horizontal component. The horizontal component of the force moves the mass a displacement of d and does work. The horizontal component would be F=cos60° and W=Fdcosθ.
For an object moving in a circle what is the Work done on the object?
W=Fdcosθ For a circle the velocity vector points at a tangent out from the circle which makes a 90° angle with the circle. So W=Fdcosθ=0
What is the definition of work?
W=fd=Fcosθ
What's the difference btw mass and weight?
When in the presence of a gravitational force (earth's gravity) the object has weight which is mass x gravity. In space their is no gravity but the object still has mass. Mass is intrinsic to the object no matter gravity Weight is the object's mass x gravity.
In a closed system which are exchanged with the surroundings: heat/work and mass.
Work and heat are exchanged with the surroundings but not mass.
