Work
Potential & Kinetic Energy
Equation Modifiers
Energy Conservation
Power & Past Topics
100
State the work equation and define work.
W = Fd//
W = Fd cos θ
Work is the change or transfer of energy.
100
State both the gravitational potential energy and the kinetic energy equations.
PEg = mgh
KE = ½ mv2
100
Work is done to lift an object 2m. Compare the work done to lift the same object 4 m.
Twice the work
(PEg and h are directly related)
100
What is the combination of PEg and KE called?
Mechanical energy
100
Give the unit of power as well as an equivalent unit combination.
Watt
J / s
Nm / s
kgm2 / s3
200
State the units of work in two different ways.
Joule
Newton x meter
kg m2 / s2
Watt x sec
200
Describe the change of energy when a textbook is falling to the ground, and then after it impacts the ground.
PEg is changing into KE. Upon striking the ground, this KE becomes sound and heat (thermal/internal energy, or Q).
200
Work is done to accelerate an object from rest to 10 m/s. Compare the work done to accelerate the same object from rest to 20 m/s.
Four times the work
(KE and v are exponentially related)
200
What force often does work to remove energy from moving objects?
Friction
200
Describe how an applied force can cause an object to move at constant velocity.
Constant velocity means 0 net force. The applied force must be equalized my an opposing force of equal magnitude (like friction).
300
Solve for the work done by the action of lifting a 50 N dumbbell 0.5 m.
W = Fd = ΔPE = mgh
W = (50N)(0.5m)
25 J
300
Describe what is causing the change of energy when a textbook is falling towards the ground.
Gravity works to change potential energy into kinetic energy as an object free falls.
300
An object has a PEg of 100 J at a 10 m above the ground. What is its PEg if it is moved to a 5 m shelf on a planet where the acceleration is gravity is half the Earth’s?
PEg = mgh
All variables directly affect PEg. The new g and h are each half their original value, so the new PE is ¼ its old value.
¼ (100J) = 25 J
300
What is the total energy of a 1 kg ball moving at 2 m/s the moment it is 2 meters above the ground?
ET = PE + KE
ET = mgh + ½ mv2
ET = (1kg)(9.81m/s2)(2m) + ½ (1kg)(2m/s)2
21.62 J
300
Two students climb a staircase to measure their power. Student A (mass of 70 kg) climbs the stairs in 3 sec. Student B (mass of 90 kg climbs the stairs in 4 sec. Who is more powerful?
P = W / t = Fd / t = Fv
In this case W = ΔPE, so
P = ΔPE / t
PA = mAgh / tA and PB = mBgh / tB
g and h are the same for both students, but the m/t ratio is higher for student A.
A: 70kg/3sec = 23.333
B: 90kg/4sec = 22.5
400
Three hikers of equal mass (A, B, and C) climb a mountain. Each takes a different path, but they all meet at the peak. Who did the most work against gravity, and explain why.
They all do the same work. The work done is equivalent to PEg which is equal for all (same m, same g, same h).
400
An acorn falls from a tree branch 10 m to the ground.
Solve for its velocity upon striking the ground.
PEi = KEf
mghi = ½ mvf2
gh = ½ v2
(9.81m/s2)(10m) = (.5)(v)2
Vf = 14 m/s
400
An object has 100 J of kinetic energy while moving. What is its new KE if its velocity is tripled and its mass is doubled?
KE is directly related to mass and exponentially related to velocity. Tripled velocity means x9 KE, and doubling mass means x2 KE.
KEnew = 18KEold= 1800 J
400
A crane exerts 2300 J to lift a 50 kg crate at constant velocity. The crate rises 4 meters. Calculate the energy lost due to friction.
W = ΔPE + Q = 2300 J
2300 J = mgh + Q
2300J = 50kg(9.81m/s2)4m + Q
2300 J = 1962 J + Q
Q = 338 J
400
State the equation for the centripetal force and identify its direction. Also describe the relationship between the direction of centripetal acceleration and tangential velocity.
Fc = mv2 / r
Fc is always directed towards the center of a circular path.
ac and vtangential always act at a right angle to each other, with ac directed towards the center of the circular path
500
A mover pushes an 80 N box with a force of 40 N up a 3 m long incline onto a truck. The truck is 1 meter off the ground. Calculate the work done by the mover.
W = Fd = 40N(3m) = 120 Joules.
While the gain in PE is only 80 J here, the mover’s work has to give the box this much energy plus overcome friction. This means friction took and extra 40 J to overcome.
500
Calculate the work done by an engine to accelerate a 2000 kg car from 5 m/s to 15 m/s.
W = ΔKE = ½ m(Δv)2
W = ½ (2000kg)(10m/s)2
W = 100,000 J
500
A train moving at 20 m/s needs 400 m to stop when the emergency brake is pulled. How much distance would it need to stop if it were moving at 40 m/s?
Double v → x4 KE → x4 Work = x4 Fd → 1600 m
Velocity is exponentially related to KE. KE (Work) is directly related to distance and force, but force, (and acceleration) were constant in both cases.
500
A car engine works to exert 10,000 J of energy to accelerate a 2000 kg car from rest. The car moves at 3 m/s as a result. Calculate energy lost due to friction.
1,000 J
W = KE + Q = ½ mv2 + Q
KE = ½ 2000kg(3m/s)2 = 9,000J
If 10,000 J of work was done, then 1,000 J was needed to overcome friction.
500
A baseball and a bowling ball are rolled off the same desk and fall to the floor. The baseball was rolling faster than the bowling ball. Which object takes more time to fall to the ground? Explain.
Both objects land in the same amount of time. They fall from the same height, the both have viy = 0, and they both have the same acceleration (g).