Work and Energy Practice Questions
1. Calculate the work done in each of the scenarios below.
a) A person uses 1.00 x 102 N to move a 15.0 kg object 5.0 m along the floor
at a constant speed.
b) 1.00 x 102 N of force is used at an angle of 30.0o to the horizontal to
move a 15.0 kg object 5.0 m across a floor at a constant speed.
c) A 15.0 kg object is moved straight up 5.0 m at a constant speed.
2. A person carries a 200.0 N package up a flight of stairs that has a height
of 10.0 m. The horizontal force needed for these stairs was 50.0 N for a
horizontal distance of 35.0 m. How much work does the person do on the
package during this entire motion?
3. A force of 50.0 N acts on the block at 30.0o to the horizontal, moving it a
horizontal distance of 3.0 m. How much work is done?
4. How much work is done by an applied force to lift a 15-Newton block 3.0
meters vertically at a constant speed?
5. A person with a mass of 80.0 kg runs at a constant speed up some stairs
that have a vertical height of 8.0 m. Determine the amount of vertical work
done.
6. Calculate the work done by a 2.00 N force directed 30.0° to the vertical
to move a 5.00 x 102 g box a horizontal distance of 4.00 x 102 cm across a
floor at a constant speed of 0.5 m/s. (make sure to convert any necessary
units.)
7. A 3.25 x 102 kg metal box is dragged across a flat cement floor by a chain
aligned at an angle of 65.0o to the vertical. The frictional coefficient of the
floor is 0.450. What should be the magnitude of F so that the net work
done by it and the kinetic frictional force is zero?
8. A starship is travelling through space at a velocity of 2.25 x 104 m/s and
has a mass of 6.75 x 104 kg. The thrusters push the starship along with a
constant force of 3.18 x 105 N, causing a displacement of 2.50 x 106 m.
What is the final velocity of the starship?
9. The hammer throw is a track and field event in which a 7.3 kg ball (the
hammer), starting from rest, is whirled around in a circle several times and
released. It then moves upward on the familiar curving path of projectile
motion. In one throw, the hammer is given a speed of 29 m/s. Determine
the work done to launch the motion of the hammer.
10. A 0.075 kg arrow is fired horizontally. The bowstring exerts an average
force of 65 N on the arrow over a distance of 0.90 m. With what speed does
the arrow leave the bow?
11. A 65.0 kg jogger is running at a speed of 5.30 m/s.
a) What is the kinetic energy of the jogger?
b) How much work is done by the net external force that accelerates the
jogger to 5.30 m/s from rest?
12. The speed of a hockey puck decreases from 45.00 m/s to 44.67 m/s in
coasting 16 m across the surface of the ice. Find the coefficient of kinetic
friction between the puck and ice.
13. A cart is loaded with and object and pulled at constant speed along an
inclined plane to the top. If the mass of the loaded cart is 3.0 kg and the
height reached is 0.45 m, what is the potential energy of the loaded cart?
14. If a force of 15.0 N is used to drag the loaded cart from previous
question along the incline for a distance of 0.90 m, how much work is done
on the loaded cart?
15. Determine the kinetic energy of a 1000.0 kg roller coaster car that is
moving with a speed of 20.0 m/s.
16. If the roller coaster car in the above problem were moving with twice
the speed, then what would be its new kinetic energy?
17. A 50.0 kg diver jumping off a platform had a kinetic energy of
1.50 x 104 J just before hitting the water. What is the diver's speed?
18. A 7.50 x 102 kg compact car moving at 100.0 km/hr has approximately
2.90 x 105 J of kinetic energy. What is the kinetic energy of the same car if it
is moving at 50.0 km/hr? (Either convert km/hr to m/s and use the formula,
or try and use proportional reasoning to find your answer)
19. Two people are in the weightlifting room. Ken lifts a 45.0 kg barbell over
his head 20 times in 60.0 s. Doug lifts the same barbell over his head 20
times in 10.0 s.
a) Which person does the most work?
b) Which person delivers the most power?
20. A work-energy theorem question. A slingshot fires a pebble from the
top of a building at a speed of 10.0 m/s. The building is 20.0 m tall. Ignoring
air resistance, find the speed with which the pebble strikes the ground
when the pebble is fired:
a) horizontally
b) vertically straight down
c) vertically straight up
21. Michael has double the mass of Albert. Both run up a ramp, but Albert
reaches the top of the ramp in half the time.
a) Which person did the most work?
b) Which person showed the most power?
22. A person lifts their 60.0 kg body a distance of 0.350 m in 2.00 s. How
much power is this person using?
23. An average of 30 passengers, each of which is about 70.0 kg in mass,
use an escalator to go to the second floor in a department store every 2.00
minutes. If the second floor is 4.50 m above the main floor, how much
power does the escalator use?
24.In physics, how do we define work?
25. At an angle of 40.0o to the horizontal, a 300.0 N force is applied to an
object to move a 15.0 kg object at a constant speed for 20.0 m. How much
work is done?
26. What is the definition of potential energy? Name two types of potential
energy
27. What is the potential energy of a 75 N object that is carried up a ladder
from the ground to a height of 15 m?
28. Define kinetic energy.
29. What is the approximate kinetic energy of a 5.0 kg bowling ball rolling
along at 2.5 m/s?
30. Name three types of quantities in physics that are considered to be
vector quantities.
31. Define mechanical energy.
32. What power is spent to lift a brick when 800.0 J is used to lift the brick
3.0 m in 5.0 seconds?
33. A person lifts their 80.0 kg body up to a bar while doing a chin-up. If this
person raised their body up a distance of 45 cm in a time of 3.0 s, how
much power was being used?
34. A person is on a toboggan sliding across a patch of ice. The velocity of
the toboggan decreases from 3.10 m/s to 1.82 m/s in coasting 12.5 m
across the surface of the ice. Find the coefficient of kinetic friction between
the toboggan and ice.
35. Use the work-energy theorem to solve this question. A trebuchet is
loaded with a boulder and mounted at the top of a 45.0 m cliff. It's about to
fire when it malfunctions and fires the boulder nearly straight up into the
air, just clearing the cliff edge and falling to the ground below. It shot the
boulder up with a speed of 17.2 m/s. With what speed does the boulder hit
the ground?
36. A skid of pallets has a mass of 9.36 x 103 kg. A tractor drags it across a
floor by a chain aligned at an angle of 15.0o to the horizontal. If the
frictional coefficient of the floor is 0.279, with what force should the tractor
pull so that net work done by it and the kinetic frictional force is zero?
37. A floor has a coefficient of friction of 0.755. A rope set up at 70.0o to the
vertical is tied to a large crate and is being pulled across the floor. If the
crate has a mass of 428 kg, with how much force should the crate be pulled
if you want to pull it yet not accelerate it?
Solutions
1.
a) W = (1.00 x 102 N)(5.0 m)(cos0o) = 5.0 x 102 J
b) W = (1.00 x 102 N)(5.0 m)(cos30.0o) = 4.3 x 102 J
c) W = (150 N)(5.0 m)(cos0o) = 7.5 x 102 J, The applied force must be 150 N
since the 15 kg mass is lifted at constant speed against gravity, and since F
and d are in the same direction the angle is 0 degrees.
2.
For the vertical part, W = (200.0 N)(10.0 m)(cos0o) = 2.00 x 103 J
For the horizontal part, W = (50.0 N)(35.0 m)(cos0o) = 1.75 x 103 J
So total work done is 2.00 x 103 J + 1.75 x 103 J = 3.75 x 102 J
3.
W = (50.0 N)(3.0 m)(cos30o) = 1.3 x 102 J
4.
W = (15 N)(3.0 m)(cos0) = 45 J
5.
W = (800 N)(8.0 m)(cos0.0o) = 6.4 x 103 J
6.
Recall that the angle theta in the work equation is the angle between F and
d. In this problem, the d is horizontal and the F is at a 60 degree angle to
the horizontal. So theta = 60 degrees. So W = (2.0 N)(4.00 m)(cos60.0o) =
4.00 J
7.
WF + Wf = 0
WF = FdcosƟ
WF = Fdcos25.0o
WF = Fd(0.906)
WF = 0.906Fd
FN + FsinƟ - Fg = 0
FN + Fsin25.0o - mg = 0
FN + Fsin25.0o - (3.25 x 102 kg)(10 m/s2) = 0
FN + F(0.422) - 3250 = 0
FN + 0.422F - 3250 = 0
FN = -0.422F + 3250
Ff = uFN
Ff = (0.450)(-0.422F + 3250)
Ff = -0.189F + 1462
Wf = -Ffd
Wf = -(-0.189F + 1462)d
Wf = (+0.189F - 1462)d
Wf = 0.189Fd - 1462d
WF + Wf = 0
0.906Fd + 0.189Fd - 1462d = 0
0.906F + 0.189F - 1462 = 0
1.095F -1462 = 0
1.095F = 1462
1.095F = 1462
1.095 1.095
F = 1335 N or 1.34 x 103 N
8.
W = KEf - KEi
KEf = W + KEi
Find each part separately:
W = Fd
W = (3.18 x 105 N)(2.50 x 106 m)
W = 7.95 x 1011 J
KEi = ½mvi2
KEi = 0.5(6.75 x 104 kg)(2.25 x 104 m/s)2
KEi = 1.71 x 1013 J
KEf = W + KEi
KEf = 7.95 x 1011 J + 1.71 x 1013 J
KEf = 1.79 x 1013 J
To find the final velocity, use:
KEf = ½mvf2
1.79 x 1013 J = 0.5(6.75 x 104 kg)v2
1.79 x 1013 J = (3.37 x 104 kg)v2
1.79 x 1013 J = (3.37 x 104 kg)v2
3.37 x 104 kg 3.37 x 104 kg
5.31 x 108 = v2
v = 2.30 x 104 m/s
9.
W = KEf - KEi
W = ½mvf2 - ½mvi2
W = ½(7.3 kg)(29 m/s)2 - ½(7.3 kg)(0)2
W = 3.1 x 103 J
10.
We'll need W = ½mvf2 - ½mvi2 and W = Fd for this question.
Since vi = 0, W = ½mvf2 - ½mvi2 right away becomes W = ½mvf2.
And since we're given F and d, we can put these in place of W because
W = Fd, and our formula becomes:
W = ½mvf2
Fd = ½mvf2
(65 N)(0.90 m) = ½(0.075 kg)vf2
58.5 = (0.0375)vf2
58.5
= (0.0375)vf2
0.0375
0.0375
2
1560 = vf
vf = 3.9 x 101 m/s
11. a)
KE = ½mv2
KE = ½(65.0 kg)(5.30 m/s)2
KE = 913 J
b) Because W = KEf - KEi, and because the jogger started from rest and
therefore had no initial KE, W = KE.
W = 913 J
12.
The work done in slowing down the puck is:
W = ½mvf2 - ½mvi2
But because we're being asked for the coefficient of friction, you know we'll
have to somehow include :
Ff = uFN.
We don't know the normal force, but we do at least know what gravity is.
Therefore, change this to:
Ff = umg
It is the friction that does the work. Also the force of friction is opposite to
the direction of motion and therefore needs to be negative:
W = -Ffd
W = -umgd
Lastly, since W = ½mvf2 - ½mvi2 and W = -umgd, we can write:
½mvf2 - ½mvi2 = -umgd
½vf2 - ½vi2 = -ugd
Now solve for u:
½(44.67 m/s)2 - ½(45.00 m/s)2 = -u(10 m/s2)(16 m)
997.7 - 1012.5 = -u(160)
-14.8 = -u(160)
-14.8 = -u(160)
-160
-160
u = 0.0925 or 9.3 x 10-2
13.
PE = mgh
PE = (3.0 kg)(10 m/s2)(0.45 m)
PE = 13.5 J or 14 J
14.
W = FdcosӨ
W = (15 N)(0.90 m)(cos0o)
W = 13.5 J or 14 J
15.
KE = ½mv2
KE = ½(1000.0 kg)(20.0 m/s)2
KE = 200 000 J or 2.00 x 105 J
16.
If the speed is doubled, the KE is quadrupled, thus
KE = 4(2.00 x 105 J)
KE = 8.00 x 105 J
or
KE = ½(1000.0 kg)(40.0 m/s)2
KE = 8.00 x 105 J
17.
KE = ½mv2
1.50 x 104 J = ½(50.0 kg)v2
1.50 x 104 = (25.0)v2
1.50 x 104 = (25.0)v2
25.0
25.0
2
600 = v
v = 24.5 m/s
18.
The KE is directly related to the square of the velocity. If the velocity is
reduced by a factor of 2, then the KE will be reduced by a factor of 4. Thus
the new KE is 2.90 x 105 J/4 = 7.25 x 104 J.
19.
a) Both do the same work, since both perform the same Fd.
b) Doug. Both might do the same work, but Doug does the work faster.
20.
a) Recall that a projectile shot straight out horizontally and a projectile
dropped would, provided they were released simultaneously, hit the
ground at the same time. Therefore vi for part a) is 0.00 m/s.
(KEi + PEi) = (KEf + PEf)
½mvi2 + mghi = ½mvf2 + mghf
½vi2 + ghi = ½vf2 + ghf
½(0.00 m/s)2 + (10 m/s2)(20.0 m) = ½vf2 + (10 m/s2)(0.00 m)
200.0 = ½vf2
400.0 = vf2
vf = 20.0 m/s
b)
½vi2 + ghi = ½vf2 + ghf
½(10.0 m/s)2 + (10 m/s2)(20.0 m) = ½vf2 + (10 m/s2)(0.00 m)
50.0 + 200.0 = ½vf2
250.0 = ½vf2
500.0 = vf2
vf = 22.4 m/s
c) still 22.4 m/s
21.
a) Michael, because he must use twice the force to lift his twice as massive
body up the ramp.
b) Both have the same power. Albert might have ran up the ramp in half
the time, but he is also only half as massive.
22.
P = Fd/t
P = (600 N)(0.350 m)/2.00 s
P = 105 W
23.
P = Fd/t
P = (30)(700 N)(4.50 m)/120 s
P = 787.5 W or 7.88 x 102 W
24.
A force acting upon an object to cause a displacement
25.
W = FdcosӨ
W = (300.0 N)(20.0 m)(cos40.0o)
W = 4596 J or 4.60 x 103 J
26.
Stored energy of position. Elastic and gravitational potential energy.
27.
PEgrav = mgh
PE = (75 N)(15 m) *careful... 75 N is the mg already combined for you
PE = 1125 J or 1.1 x 103 J
28.
Energy of motion.
29.
KE = 0.5mv2
KE = (0.5)(5.0 kg)(2.5 m/s)2
KE = 15.625 J or 1.6 x 101 J
30.
Velocity, force, acceleration, among others. Any quantity that has an
amount and a direction.
31.
The energy acquired by the objects upon which work is done, or the energy
which is possessed by an object due to its motion or its stored energy of
position. E = KE + PE
32.
P = W/t
P = 800.0 J/5.0 s
P = 160 W or 1.6 x 102 W
33.
P = Fd/t
P = (800 N)(0.45 m)/3.0 s
P = 106.7 W or 1.1 x 102 W
34.
½vf2 - ½vi2 = -ugd
½(1.82)2 - ½(3.10)2 = -u(10)(12.5)
1.65 - 4.80 = -u(125)
-3.15 = -u(125)
u = 2.52 x 10-2
35.
(KEi + PEi) = (KEf + PEf)
½mvi2 + mghi = ½mvf2 + mghf
½vi2 + ghi = ½vf2 + ghf
½(17.2 m/s)2 + (10 m/s2)(45.0 m) = ½vf2 + (10 m/s2)(0.00 m)
147.92 + 450 = ½vf2
597.92 = ½vf2
1195.84 = vf2
vf = 3.46 x 101 m/s
36.
WF + Wf = 0
WF = FdcosƟ
WF = Fdcos15.0o
WF = Fd(0.965)
WF = 0.965Fd
FN + FsinƟ - Fg = 0
FN + Fsin15.0o - mg = 0
FN + Fsin15.0o - (9.36 x 103 kg)(10 m/s2) = 0
FN + F(0.258) - 9.36 x 104 = 0
FN + 0.258F - 9.36 x 104 = 0
FN = -0.258F + 9.36 x 104
Ff = uFN
Ff = (0.279)(-0.258F + 9.36 x 104 )
Ff = -0.0719F + 2.61 x 104
Wf = -Ffd
Wf = -(-0.0719F + 2.61 x 104)d
Wf = (+0.0719F - 2.61 x 104)d
Wf = +0.0719Fd - 2.61 x 104d
WF + Wf = 0
0.965Fd + 0.0719Fd - 2.61 x 104d = 0
0.965F + 0.0719F - 2.61 x 104 = 0
1.0369F = 2.61 x 104
F = 2.52 x 104 N
37.
WF + Wf = 0
WF = FdcosƟ
WF = Fdcos20.0o
WF = Fd(0.939)
WF = 0.939Fd
FN + FsinƟ - Fg = 0
FN + Fsin20.0o - mg = 0
FN + Fsin20.0o - (428 kg)(10 m/s2) = 0
FN + F(0.342) - 4280 = 0
FN + 0.342F - 4280 = 0
FN = -0.342F + 4280
Ff = uFN
Ff = (0.755)(-0.342F + 4280)
Ff = -0.258F + 3231
Wf = -Ffd
Wf = -(-0.258F + 3231)d
Wf = (+0.258F - 3231)d
Wf = +0.258Fd - 3231d
WF + Wf = 0
0.939Fd + 0.258Fd - 3231d = 0
0.939F + 0.258F - 3231 = 0
1.197F = 3231
F = 2.70 x 103 N