Work, Energy, and Power Review
Physics
1. A helicopter (m = 1250 kg) is cruising at a speed of 25.0 m/s at an altitude of 185
m. What is the total mechanical energy of the helicopter?
E = KE + PE
E = ½ mv2 + mgh
E = ½ (1250)(25)2 + (1250)(9.8)(185)
E = 2.66x106 J
2. A 75 kg crate is pulled across a horizontal floor by a 350 N force, which is
directed 35 above the horizontal. The force of friction on the crate by the floor is
125 N. As the crate is moved 7.0m, what is the work done on the crate by:
a. The pulling force?
b. The gravitational force?
c. The normal force?
d. The frictional force?
e. What is the total work done on the crate?
a. WP = FPdcos
WP = (350N)(7.0)cos(35)
WP = 2006.9 J
b. Wg = Fgdcos
Wg = (75 kg)(9.8m/s2)(7.0)cos90
Wg = 0 J
c. WN = FNdcos
WN = FNdcos90
WN = 0 J
d. Wf = Ffdcos
Wf = (125 N)(7)cos180
Wf = -875 J
e. W = WP + Wg + WN + Wf = 2006.9 J + 0 J + 0 J + -875 J = 1132 J
(Remember: Work is only done by forces acting along the line of motion.)
3. Larry's gravitational potential energy is 1870 J as he sits 2.20 m above the ground
in a sky diving airplane. What is Larry's gravitational potential energy when be
begins to jump from the airplane at an altitude of 923 m? How fast is he going
when he hits the ground?
PE = mgh
PE = mgh
1870 J = m (9.8)(2.2 m)
PE = (86.7)(9.8)(923 m)
m = 86.7 kg
PE = 784236.2 J = 7.8x105 J
KE = ½ mv2
784236.2 = ½ (86.7)v2
v = 134.5 m/s
4. A pebble rolls off the roof of the high school and falls vertically. Just before it
reaches the ground, the pebble's speed is 17 m/s. Neglect air resistance and
determine the height of the high school.
E0 = E1
KE0 + PE0 = KE1 + PE1
½ mv02 + mgh0 = ½ mv12 + mgh1
mgh0 = ½ mv12
gh0 = ½ v12
(9.8)h0 = ½ (17)2
h0 = 14.7 m
5. A bicyclist is traveling at a speed of 17 m/s as he approaches the bottom of a hill.
He decides to coast up the hill and stops upon reaching the top. Determine the
vertical height of the hill.
E0 = E1
KE0 + PE0 = KE1 + PE1
½ mv02 + mgh0 = ½ mv12 + mgh1
½ mv02 = mgh1
½ v02 = gh1
½ (17)2 = (9.8)h1
h1 = 14.7 m
6.
A roller coaster starts from rest at the top of an 18-m hill as shown. The car
travels to the bottom of the hill and continues up the next hill that is 10.0 m high.
How fast is the car moving at the top of the 10.0-m hill if friction is ignored?
E0 = E1
KE0 + PE0 = KE1 + PE1
½ mv02 + mgh0 = ½ mv12 + mgh1
mgh0 = ½ mv12 + mgh1
gh0 = ½ v12 + gh1
(9.8)(18) = ½ v12 + (9.8)(10)
(9.8)(8) = ½ v12
v1= 12.5 m/s
7. An engineer is asked to design a playground slide such that the speed a child
reaches at the bottom does not exceed 6.0 m/s. Determine the maximum height
that the slide can be.
E0 = E1
KE0 + PE0 = KE1 + PE1
½ mv02 + mgh0 = ½ mv12 + mgh1
mgh0 = ½ mv12
gh0 = ½ v12
(9.8)h0 = ½ (6)2
h0 = 1.8 m
8. A car is traveling at 7.0 m/s when the driver applies the brakes. The car moves
1.5 m before it comes to a complete stop. (m = 1000 kg)
a. What is the work done on the car?
b. What is the braking force needed to bring the car to a stop?
a. W = KE = ½ mv12 – ½ mv02
W = ½ (1200 kg)(0 m/s)2 – ½ (1000 kg) (7 m/s)2
W = -24500 J
b. W = Fd
-24500 J = F(1.5 m)
F = -16333 N
9. How much power is needed to lift a 75-kg student vertically upward at a constant
speed of 0.33 m/s?
P = Fv
P = mgv
P = (75)(9.8)(0.33)
P = 242.6 W
Use the following pulley to answer #10-12:
4m
2m
20 N
10. What is the ideal mechanical advantage for this pulley? What is another way you
can find it?
IMA = din/dout = (4 m)/(2 m) = 2
You can also find IMA by doubling the number of movable pulleys or by
counting the strings.
11. What is the actual mechanical advantage of this pulley if the applied force is 12N?
AMA = Fout/Fin = (20 N)/(12 N) = 1.667
12. What is the efficiency for this pulley?
Efficiency = (AMA/IMA)*100
Efficiency = (1.667/2)*100
Efficiency = 83.3 %
OR
Efficiency = (Wout/Win)*100
Efficiency = ((20)(2)/(12)(4))*100
Efficiency = 83.3 %
13. For the lever shown below, what is the distance that the 150 kg weight will move
when there is an applied force of 15 kg over a distance of 3 m?
Wout = Win
(150)(9.8)(d) = (15)(9.8)(3)
d = 0.3 m
14. In the absence of friction, a 100 kg skier will have an acceleration of 6.93 m/s2 on
a 45 Black Diamond 1414 ft long slope. If the same skier, skis down a 15
Green Circle 3864 ft long slope, the acceleration will be 2.54 m/s2. If both skiers
start from the top of Whitetail resort (height of 1000 ft), show which skier will be
going faster when they get to the lodge at the bottom.
E0 = E1
1000 ft = 304.8 m
mgh = ½ mv2
1414 ft = 431 m
(100 kg) (9.8) (304.8 m) = ½ (100 kg) v2
3864 ft = 1177.7 m
v =77.3 m/s
(for both since the both start at the same
height)
1000 ft
45
15
15. Consider the same skier and slopes from question 15. If there is a frictional force
of 600 N between the skier and the Black Diamond (431 m long) slope. What is
the energy lost due to friction during his trip down the slope?
Wlost = Ffdcos = (600 N)(431 m) (cos0) = 258, 600 J
16. The total energy lost due to friction for the Bunny slope skier was 293, 000 J.
Find their new velocities when the get to the lodge!!
Black Diamond:
E0 - Elost= E1
mgh - Elost = ½ mv2
(100 kg) (9.8) (304.8 m) – (258,600) = ½ (100 kg) v2
v =28.3 m/s
Bunny Slope
E0 - Elost= E1
mgh - Elost = ½ mv2
(100 kg) (9.8) (304.8 m) – (293,000) = ½ (100 kg) v2
v =10.7 m/s