Physics 2204
Unit 3: Energy, Work and Power
Worksheet #11: Work - Energy Theorem
Work- Energy Theorem :
A resultant force changes the velocity of an
object and does work on that object.
PART A: Multiple Choice
1.
What is the ability to do work called?
(A)
(B)
(C)
(D)
2.
The Work-Energy Theorem states that the
(A)
(B)
(C)
(D)
3.
kg m/s
kg m/s2
kg/m/s2
kg m2/s2
A 950 kg car accelerates from rest to 7.0 m/s. How much work is done on the car?
(A)
(B)
(C)
(D)
6.
ME i = ME f
ME i = W
W = ÄPE
W = Ä KE
The SI unit of work can be expressed as:
(A)
(B)
(C)
(D)
5.
work done equals the change in the net force
work done equals the change in kinetic energy
work done equals the net force divided by the net kinetic energy
work done equals the product of the net force and the net kinetic energy
Which of the following equations expresses the work-kinetic energy theorem?
(A)
(B)
(C)
(D)
4.
energy
force
momentum
power
3300 J
6700 J
23000 J
47000 J
A bullet with a kinetic energy of 400 J strikes a wooden block where a 8.00x103 N resistive
force stops the bullet. The penetration of the bullet in the block is:
(A)
(B)
(C)
(D)
0.050 m
0.500 m
0.200 m
2.00 m
1
7.
How much energy is required to stop a car of mass 100.0 kg moving
at a speed of 25.0 m/s?
(A)
(B)
(C)
(D)
(E)
8.
A friend slides a glass of milk with a mass of 500 g along the top of a table. The coefficient of
kinetic friction between the glass and the bar is ìk = 0.10. When the glass leaves your friend’s
hand, it is moving at a speed of v0 = 2 m/s. How far does the glass travel before coming to a
stop? Take g= 9.8 m/s2.
(A)
(B)
(C)
(D)
9.
1.2 N
11 N
392 N
627 N
What average net force need be exerted on a 2 kg object traveling at 6m=s to bring it to
rest (i.e. stop it) in a distance of 3 m?
(A)
(B)
(C)
(D)
12.
47.5 N
1400 N
2060 N
19 600 N
A javelin with a mass of 1.2 kg is thrown at a speed of 28 m/s. How much force
does the athlete exert over a throwing distance of 1.5 m?
(A)
(B)
(C)
(D)
11.
1.0 m
1.8 m
2.0 m
2.2 m
A 500 kg car is moving at 28 m/s. The driver sees a barrier ahead . If the car takes 95 m to
come to rest, What is the magnitude of the minium average force necessary to stop?
(A)
(B)
(C)
(D)
10.
1150J
21,150 J
31,250 J
32,250 J
42,250 J
1N
2N
6N
12N
A car moving at 50 km/hr skids 20 m with locked brakes. How far will the car skid with locked
brakes if it is traveling at 150 km/hr?
(A)
(B)
(C)
(D)
40 m
60 m
90 m
180 m
2
PART B: Written Response
1.
Evaluate the following:
a.
b.
c.
A 2000 kg car is moving at 10.0 m/s. Calculate its KE. Answer is 100 000 J
Suppose that the speed of the car in part (a) doubled to 20.0 m/s. What would be the
new value of its KE? Answer is 400 000 J
Explain how you can use the answer from part (a) to get the answer to part (b) without
using the formula for KE. Answer :the speed means four times
2.
Do the same for the following:
a.
A 900 kg moose is running at 2.0 m/s. Calculate its KE. Answer is 1800 J
b.
Suppose that the moose triples its speed to 6.0 m/s. Calculate the new value for the KE
without using the formula. Answer is 16200 J
3.
A ball with a mass of 0.40 kg is estimated to have a KE equal to 80.0 J. Calculate its speed.
Answer is 20 m/s
4.
The KE of a speeding bullet is estimated to be 2240 J. If the mass is 0.028 kg, what is the speed?
Answer is 400 m/s
5.
A 75 kg cyclist, on a 5 kg bicycle speeds up from 10 m/s to 20 m/s.
a.
What is the kinetic energy before speeding up? Answer is 4000 J
b.
What is the kinetic energy after speeding up? Answer is 16000 J
c.
Based on (a) and (b) what happened to kinetic energy as speed doubles.
Answer is 4 x increase
d.
By what factor would the kinetic energy change if the speed increased by a factor of 3?
Explain your reasoning. Answer is 9
6.
A moving object is estimated to have a KE equal to 40 J.
a.
If you increased the speed by a factor of 5 what would be the KE? Answer is 1000 J
b.
If you increased the speed by a factor of 10 what would be the KE? Answer is 4000 J
c.
If you DECREASED the speed by a factor of 2, what would be the KE? (That is, what
would be the KE if the speed was cut in half?) Answer is 10 J
7.
A hot wheels car with a mass of 0.050 kg is moving at 0.80 m/s along a track. It passes through a
battery powered launcher which increases its speed to 1.40 m/s. By what FACTOR was the
kinetic energy increased? Answer is 3.0625 (note you can SQUARE the factor by which the
speed changed)
8.
A speeding bullet is only going about 10 times as fast as you could throw it. It packs about 100
times the WHALLOP, though. Why? Increasing the speed by a factor of 10 means that the
KE is actually 100 times greater.
9.
Rolling ball has 18 J of kinetic energy and is rolling 3.0 m/s. Find its mass. Answer is 4.0 kg
10.
Calculate the kinetic energy of a 45 g golf ball travelling at:
a)
20. m/s
Answer is 9.0 J
(b)
40. m/s
Answer is 36 J
(c)
60. m/s
Answer is 81 J
11.
Calculate the amount of work done when an 1150 kg car accelerates from
2.00 m/s to 6.00 m/s. June 2009
12.
A 605 kg race car accelerates from 20.0 m/s to 60.0 m/s. June 2010
i)
ii)
Calculate the work done during the acceleration
If the car generates 582 kW of power, calculate the time it took to accelerate.
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