CP PhysicsName:
Review- Work, Power, and Energy
USE A SEPARATE SHEET OF PAPER TO ANSWER QUESTIONS/SOLVE THE PROBLEMS!!!
Concepts
1) Write the equation for each of the following. Define each variable and list the correct units.
a) Work W = Fdcosθ, W is work, unit is JoulesF is Force, unit is Newtons, θ is the angle of the force compared to the direction of movement
b) Power P= W/t or P= Fd/t P is power, unit is J/s or Watts t is time, unit is seconds
c) Kinetic energy KE = 1/2mv2 KE is kinetic energy or energy of movement, unit is J m is mass unit kg, v is velocity unit m/s
d) Potential energyPE = mgh PE is potential energy, usually Gravitational Potential Energy of position, unit is J. m is mass, unit kg, h is height unit meters.
2)What is meant by the conservation of energy? Give an example of a situation where energy is conserved.Conservation of Energy Is the idea that energy cannot be created or destroyed, only changed from one type of energy to another. Thus, the total energy amount is the same for a cow standing at the top of the cliff (PE is say, 5000 J, KE is 0) as the cow halfway down (PE 2500 J, KE 2500 J) as the cow right at the bottom (PE 0J, KE 2500J).
3)How do potential, kinetic, and total energy change for a falling object?For a falling object, total energy is the sum of PE + KE (+ Heat energy if you count friction) So, TE = PE + KE. At the top of the path, all of the energy is PE, none is KE and none is HE, so PE = TE at the top. As it falls, some energy is used for friction with the air, but most will be converted to KE, so at the bottom TE ≈ KE and PEtop ≈ KEbottom ≈ TE
4)What is the work-energy theorem? Write the equation(s) for it. Give an example of at situation using this theorem.The work energy theory says that the amount of work you do on an object equals the amount of energy you added to it. W = ΔE For example, if I do work on a box equal to 15 J, now the box has 15 J more energy.
Problems
5)A force is applied to push a file cabinet 18 m across the floor. The work done is 1400 J.
a) How much force was exerted to move the file cabinet?
F = 77.78 N
b) If it was moved in 15 seconds, how much power was used to move the file cabinet?
P = 93.33 W or J/s
6)Oliver exerts 200 N of force in pushing a box of CDs across the floor. He pushes at an angle of 35
degrees to the horizontal. How much work is done in moving the box 10 m?
W = 1638.30 J
7)A high diver has 1200 J of potential energy. She has a mass of 60 kg.
a) How high is the diving platform?
h = 2.04 m
b) How much kinetic energy does she have halfway to the water?
PE = mgh
PE = 60x9.8x1.02 = 600 J
PE + KE = TE
600 +KE = 1200
KE = 600J
c) How fast is she going halfway to the water?
v = 4.47 m/s
8)What is the mass of a cow moving at a velocity of 0.5 m/s with 25 J of kinetic energy?
m = 200 kg
9)A 1.2 kg car is traveling at a velocity of 1.5 m/s. It is accelerated to 3.5 m/s. Calculate the change in
kinetic energy.KE = ½ mv2KE = ½ mv2
KE = ½ x 1.2 x 0.52KE = ½ x 1.2 x 0.52
KE = 1.35 JKE = 7.36 J 7.36 – 1.35 = 6.01 J
10)A crane does 3.00 x 104 J of work on a crate in order to lift it 20.0 m to the roof of a construction site
a) What is the potential energy of the crate with respect to the ground?
WE theory => W =ΔPE so PE = 3.0 x 104J
b) What is the mass of the crate?
m = 153.06 kg
11)The “Nitro” roller coaster at six flags great adventure has a chain that brings 5 cars with passengers that have a combined mass of about 8000 kg to a height of 70 meters. a) How much work is done by the chain to bring the passengers to that amazing height?
W = 5488000J (Fg = mg so Fg = 78400 N)
b) How much power is used if the coaster takes 15 seconds to get to the top of the hill?
P = 365866 J
c) How much potential energy does the coaster have at the top of that first hill?
PE = 5488000 J
d) How fast is it going if the coaster drops 35 meters on a frictionless track?
35 m is half of 70 m, so KE is half of 5488000 = 2744000 J
v = 26.19 m/s
12)A basketball that never loses energy, with a mass of 1kg is dropped from a height of 1.8 meters. a) What is the potential, kinetic and total energy just before it is dropped from that height?
PE = mgh= 17.64 J
KE = 0J
TE = 17.64 +0 = 17.64 J
b) What is the potential, kinetic and total energy at a height of 0.9 m?
PE = 8.82 J
KE = 8.82 J
TE = 17.64
c) What is the potential, kinetic and total energy at a height of 0.45 m?
PE = 4.41 J
KE = 13.23 J TE = 17.64 J
13)Describe, in complete sentences, the types of energy at points A, B, and C in the diagram above. Assume that the rollercoaster starts at rest at point A and the roller coaster does not experience any friction. Once the energies are described at all 3 points, explain how you know that the energy exists.
A)At point A, the roller coaster has all potential energy (PE = mgh = 500 x 9.8 x 47 = 230300 J) because it is up high, and no kinetic, because the cars are not yet moving, so the total is KE + PE = TE = 0 + 230300 = 230300J.
B) At point B, the total remains the same, and no energy has been lost to friction. The car has changed all its PE into KE by the time it reaches the bottom, so KE = 230300 J, PE = 0. The v is KE = ½ mv2
230300 = ½ 500 v2
v = 30.35 m/s
C)At point C, the car has begun climbing back up, so some, not all, of the KE from point B has changed back to PE. The total is still the same. PE= mgh = 500 x 9.8 x 23 = 112700 J, and TE = PE + KE so 230300 = 112700 + KE so KE = 117600 J. IF KE = 117600 then v is:
KE = ½ mv2
117600 = ½ 500 v2
21.69 m/s
v has slowed from point B to point C.