Download Ideal Mechanical Advantage

South Pasadena  A.P. Physics
Name____________________ Period___
6  Work and Energy
Practice Test
1Work
The SI unit for work is the joule (J), which equals one
Newton-meter (N-m). For maximum work to be
done, the object must move in the direction of the
force. If the object is moving at an angle to the force,
determine the component of the force in the direction
of motion, using
3. After finishing her physics homework, Sarah pulls
her 50.0 kg body out of the living room chair and
climbs up the 5.0 m high flight of stairs to her
bedroom. How much work does Sarah do in
ascending the stairs?
(Hint: The number of stairs or the slope of the stairs
is irrelevant. All that is important is the change in
position.)
W = F x displacement cos θ
Remember, if the object does not move, or moves
perpendicular to the direction of the force, no work has
been done.
Problems:
1. Bud, a very large man of mass 130 kg, is pulling on
the rope attached to the crate with a force of 450 N.
He pulls at an angle of 38  as shown. There is a
frictional force of 125 N.
a) If the crate moves a distance of 55 cm, how much
work does Bud do on the crate?
b) If the crate has a mass of 65 kg, what would be
the acceleration of the cart?
2  Power
Power is the rate at which work is done.
Power =
work
Elapsed time
The SI unit for power is the watt (W), which equals
one joule per second (J/s). One person is more
powerful than another if he or she can do more work in
a given amount of time, or can do the same amount of
work in less time.
Problems:
4. In the problem in station 1 where Sarah (50.0 kg)
climbed the 5.0 meter high staircase, she took 10.0
seconds to go from the bottom to the top. The next
evening, in a rush to catch her favorite TV show, she
runs up the stairs in 3.0 seconds.
a) What values of power does she generate each night?
2.
Fill in the missing information:
Work
(J)
Force (N)
Distance (m)
a) 45.0
?
2.0
b) 122
3.0
?
c)
?
28.0
30.0 cm
?
75.0 kg firefighter
climbs a flight of
stairs.
d)
10.0 m high
b) On which night does Sarah do more work?
c) On which night does Sarah generate more power?
3
5. Fill in the missing information:
Power (Watts)
Work (Joules)
Time
(seconds)

Potential
Energy
Energy is the ability to do work. The two kinds of
energy we are dealing with in this chapter are the
energy of position or stored energy (potential energy)
and the energy of motion (kinetic energy).
Gravitational potential energy relies upon the vertical
change in height and not upon the path taken.
a)
60.0
20.0
?
b)
240
?
2.5
c)
?
183
5.5
d) 3.5kilowatts
4.8 x 10-3
?
Potential Energy = m g h
2.0 megajoules
25.0
Other forms of stored energy exist, such as when a bow
is pulled back and before it is released, the energy in
the bow is equal to the work done to deform it. This
stored or potential energy is written as Δ PE = F Δd
The unit of potential energy (like work) is Joules.
e)
?
Problems:
6.
Which requires more power: Lifting a 2.0 kg
object to a height of 2.0 m in a time of 2.0 seconds
or lifting a 4.0 kg object to a height of 1.0 meter in
a time of 3.0 seconds? (Hint: mass in kg is not a
force)
7. Atlas and Hercules, two carnival sideshow
strong men, each lift 200.0-kg barbells 2.00 m
off the ground. Atlas lifts his barbells in 1.00 s
and Hercules lifts his in 3.00 s.
a) Which strong man does more work?
b) Calculate which man is more powerful.
8. Legend has it that Isaac Newton “discovered”
gravity when an apple fell from a tree and hit
him on the head. If a 0.20 kg apple fell 7.0 m
before hitting Newton, what was its change in
PE during the fall?
9. It is said that Galileo dropped objects off the
Leaning Tower of Pisa to determine whether heavy
or light objects fall faster. If Galileo had dropped a
5.0 kg cannon ball to the ground from a height of
12 m, what was the change in PE of the cannon
ball?
Fill in the missing information:
10.
Potential
Energy (J)
Mass (kg)
Height
(m)
a)
50.0
75.0
?
b)
280
?
1.8
c)
?
17.3
5.5
d) 3.5 kilojoules
4.8 x 10-3
?
14. A car with 54,000 joules of kinetic energy
is moving at 35 m/s? What is the car’s
mass?
15. An oxygen molecule of mass 5.31 x 10-26 kg,
has a kinetic energy of 6.21 x 10-21 Joules.
How fast is it moving?
16. If the kinetic energy of an arrow is doubled,
by what factor has its velocity increased?
4  Kinetic
Energy
KE = ½ mv2
Kinetic Energy is the energy of motion and varies with
the square of the speed. Kinetic Energy = ½ mass x
(velocity)2 and the SI unit of KE is also Joules, which
is the same unit used for work. When work is done on
an object, energy is transformed from one form to
another. The sum of the changes in potential, kinetic
and heat energy is equal to the work done on the
object.
17. If the velocity of the arrow is doubled, by
what factor does its kinetic energy increase?
5 
11. A greyhound at a race track, can run at a speed
of 16.0 m/s. What is the K.E. of a 20.0 kg
greyhound as it crosses the finish line?
12. A 7.0 kg bowling ball is moving at 2.0 m/s.
What is it’s Kinetic Energy
Conservation
Energy
of
According to the law of conservation of energy, energy
cannot be created or destroyed, but remains constant in
a system, when no forces are acting other than gravity.
Δ KE = ΔPE
or
Total Mechanical Energy = P.E. + K.E.
KE I + PE I = KE f + PE f
13. An 1800 kg truck has a kinetic energy of
95,000 Joules. What is the truck’s velocity?
For a bowling ball, the equation simplifies to mgh = ½
mv2, so then Velocity at bottom = √2gh
6 
Problems:
Work-Energy
Theorem
Work done on an object = change in its K.E.
F x d = change in Kinetic Energy
18. A 7.5 kg bowling ball is brought back to
a height of 1.2 meters and released.
How much kinetic energy will it have at
the lowest point in its swing?
19. What will be the velocity of the bowling
ball above at the lowest point?
20. When the bowling ball above is released,
how much kinetic energy will it have when
it is 0.60 meters above its lowest point?
21. How much kinetic energy will the bowling
ball have when it swings to the highest
point on the other side of its swing?
22. How much potential energy will the
bowling ball have when it swings to the
highest point on the other side of its swing?
24. How much work must be done to stop a
1000 kg car traveling at 31 m/s?
25. When the brakes of a motorcycle traveling
at 60 km/hr become locked, how much
farther will the motorcycle skid than if it
were traveling at 20 km/hr?
26. How much work is required to stop an
electron (m= 9.11 x 10-31 kg), which is
moving with a speed of 1.90 x 106 m/s?
27. A car does 7.0 x 104 J of work in traveling
2.8 km (2,800 m) at constant speed.
What was the average retarding force
(from all sources) acting on the car?
7

Efficiency
Efficiency is the ratio of the work output to the
work input and has no units and is usually
expressed as a percentage.
23. If you turn on the radio in your car with the
engine off, will more gasoline be burned
later when the car engine is turned on, as a
result of having turned on the radio?
a)
YES
b) NO
Efficiency = work output x 100
work input
or
AMA x 100
IMA
28. A bicycle rider is doing 35 joules of work
for a return of 32 joules of work output by
the bicycle. What is the efficiency of the
bicycle?
8Machines
Machines are devices that help do work by
changing the magnitude or direction of the
applied force.
32. What is the mechanical advantage of this
machine?
33. A crate of bananas weighing 3000 N is
shipped from South America to New York,
where it is unloaded by a dockworker, who
lifts the crate by pulling on the rope of a
pulley system with a force of 200 N. What
is the actual mechanical advantage of the
pulley system?
Work In = Work Out
or
Fxd = Fxd
Examples of machines include:
A lever A pulley An inclined plane
34. If the worker above lifted the crate of
bananas a distance of 10 m,
what distance of rope did he pull?
FORMULAS for MACHINES
Work in = Work out
Actual Mechanical Advantage = F out/F in
or
Ideal Mechanical Advantage =
distance in / distance out
29. If a force of 25 Newton is applied in order
to lift a 45 Newton weight a distance of
125 cm, then what distance must the applied
force be moved, using a lever?
30. What is the mechanical advantage of this
machine?
31. A lever is used to lift a 25 kilogram lead
mass a distance of 40 cm. If the distance
that the input force is moved is 80 cm, what
is the amount of force used to lift the mass?
9

Miscellaneous
35. If the dials on an electric meter read 23810
for the initial reading and then read 23890
for the final reading 3 days later, what is the
cost for the energy if the Edison company
charges $.15 per kilowatt - hour?
(Hint: Find the # of kW-hrs used by the
difference and multiply by the cost per kW-hr)
36. If you leave on a 100 watt light bulb in a
lamp for a time period of 10 hours, how
many kilowatt - hours of energy is being
consumed? (Hint: Convert watts to
kilowatts and multiply by the # of hours.)
37. How much will it cost to operate that 100
watt bulb for the 10 hours? (Hint: Multiply
your answer to # 36 by cost/kW-Hr)
38. Which is more costly? Operating a 60 watt
bulb for 5.0 hours or operating a 100 watt
bulb for a time period of 2.5 hours?
10 
Lab Questions
42. In a laboratory experiment, a student tries to
duplicate the demonstration shown in class,
using the steel ball pendulum and the razor
blade.
If the distance that the steel ball fell is 89.5
cm and it landed 69.7 cm horizontally from
the point it was cut by the razor blade, then
how many cm above the lowest point was
the steel ball brought and released?
(Hint: Calculate the kilowatt-hours for each
bulb and then multiply by $.15 per kilowatt –
hour)
39. Is a hand-held generator easier or harder to
crank when a light bulb is screwed into it?
a)
easier
b) harder
40. Is a hand-held generator easier or harder to
crank when the light bulb is screwed into it
but it is burned out?
a)
easier
razor
_____________________________
floor
Consider the lab done in class called
“Making the Grade” where a cart was pulled
up a ramp of varying angles.
b) harder
41. If two identical cars are speeding along and
then they both apply their brakes, for the car
traveling at twice the speed, how much
more distance will be required to come to a
stop, compared to the car that is traveling at
half the speed?
a) both cars require the same braking
distance
b) faster car will require twice the braking
distance
c) faster car will require four times the
braking distance.
d) faster care will require nine times the
braking distance.
43. Considering that varying angles were used
for the placement of the wooden board, was
cosine  used in the calculation of the work
done? Explain why or why not?
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44. Would you expect the calculated work to be
approximately the same or to be different
for two different trials of a 25  angle of
incline and a 45  angle of incline? Explain
why or why not?
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ANSWERS to Chapter 6 Practice
Test (Work and Energy)
6. Lifting 2.0 kg mass (20 Watts):
(Other mass requires 13 W)
1. a) Force in direction of motion
7. a) Both do the same amount of work
b) Atlas is more powerful (4,000 Watts)
= 450 N cos 38  = 354.6 N
Work = F x d = 354.6 N x 0.55 m = 195 J
or 2.0 x 102 Joules
b) Net Force = Pulling Force – Friction Force
= 354.6 N – 125 N = 230 N
a = Fn/m = 230 N / 65 kg = 3.5 m/s2
2.
8. 0.20 kg x 9.8 m/s2x 7.0 m =13.7 J
9. 5.0 kg x 9.8 m/s2x 12 m =588 J
10.
Work (J)
Force (N)
Distance (m)
45.0
22.5 N
2.0
122
3.00
40.7 m
8.40 J
28.0
30.0 cm
75.0 kg
firefighter
climbs a flight of
stairs 10.0 m high
7,350 J
3. Weight = 50.0 kg x 9.8 m/s2 = 490 N
Work = 490 N x 5.0 m = 2,450 J
4.
a) Sarah does the same work both nights
b) More power generated 2nd night
nd
c) 1st night = 250 W; 2 night = 817 W
Potential Energy
(J)
Mass (kg)
Height (m)
50.0
75.0
0.0680 m
280
15.9 kg
1.80
932 J
17.3
5.50
3.5 kilojoules
4.8 x 10-3
74,405 m
11. KE = ½ x 20.0 kg x (16.0 m/s)2 = 2,560 J
12. KE = ½ x 7.0 kg x (2.0 m/s)2 = 14 J
13. v = (2 x 95,000 J/1800 kg)1/2 = 10.3 m/s
14. mass = (2 x 54,000 J)/ (35 m/s)2 = 88 kg
15. v = (2 x 6.21 x 10-21 J/5.31 x 10 –26 kg)1/2
= 484 m/s
5.
Power (Watts)
Work (Joules)
Time
(seconds)
20.0
.33 s
240
600 J
2.5
33 W
183
5.5
3.5 kilowatts
4.8 x 10-3
1.4 x 10-6
80,000 W
2.0 megajoules
25.0
60.0
→ 2 KE = ½ mv2
v = √2 =1.41 1.41 x greater
16. KE = ½ mv2
17. KE = ½ mv2 → KE = ½ m2v2
2v2 = 4x greater
18. PE I = KE f = 7.5 kg x 9.8 m/s2 x 1.2 m
= 88 J
19. v = √(2gh) = √(2 (9.8 m/s2) 1.2 m)
= 4.849 or 4.8 m/s
33. A.M.A. = 3,000 N/200 N = 15
34. d = (3,000 N x 10 m) / 200 N =150 m
20. (K.E. – mgh)
= 88 J – (7.5 kg x 9.8 m/s2 x 0.6 m)
= 88 J – 44 J = 44 J
35. 80 kilowatt-hr x $.15 /kilowatt-hr = $12
36. 100 watt x 1 kilowatt/1,000 watts x 10 hrs.
= 1 kilowatt-hr
21. 0 J
37. 1 kilowatt-hr x $.15 /kilowatt-hr =15 ¢
22. 88 J
23. YES
38. 60 watt bulb
NO
24. Work = F x d = ΔKE
39. harder because work is being done
= ½ x 1000 kg x (31 m/s)2 = 480,500 J
25. F x d = ΔKE
(60/20)2 = 32 x further
= 9x further
40. easier because no work is being done on
the electrons
26. F x d = ΔKE
= ½ x 9.11 x 10 -31 x (1.90 x 10 6 m/s)2
= 1.64 x 10 -18 J
41. c)
27. W = F x d so F = 7.0 x 10 4 J/ 2,800 m
= 25 N
43. Cosine  was not used and was not needed
since the angle between the applied force
and the displacement of the cart was always
zero degrees.
28. Efficiency = 32 J / 35J x 100 = 91.4 %
29. F x d = F x d d = (45 N x 125 cm)/ 25 N
= 225 cm
30. M.A. = 225 cm / 125 cm = 1.8
31. F = (25 kg x 10 m/s2 x 40 cm)/ 80 cm
=125 N
32. M.A. = 80 cm/ 40 cm = 2
42. 89.5 cm
43. You would expect that the calculated values
for work would be the same for any angle,
since the elevation for each trial was
identical and therefore according to
conservation of energy, the amount of work
to bring an object to a certain height is the
same regardless of the path that is taken.