Download 2.Work and Energy Solutions

2.Work and Energy Solutions
Work and Energy
Recall: Pushing Baby Carriages (and Lawn Mowers) vs Pulling Sleds =
F
­Fy
N = µmg
N = µmg
θ
F
=
Fy
θ
Fx = Fcosθ
f
Fx
f
mg
mg
The work done on the box is equal to the applied force x displacement moved
W = Fxd = (Fcosθ)d
=
F
­Fy
N = µmg
θ
Fx
f
mg
d
The amount of work done is equal to the energy required to move the box. • When the force is parallel to the displacement, cos θ = cos 0= 1; W=Fd
> Energy is being added to the system > Work is being done ON the system
• When the force is antiparallel to the displacement, cos θ = cos 180= ­1; W=­Fd
> Energy is being removed from the system
> Work is being done BY the system
• When the force is perpendicular to the displacement, cos θ = cos 90= 0; W= 0
> Energy is neither being added, or removed, from the system
> No work is done
• Power is the amount of energy used/ spent per unit time: P = W/t= Fxd/t
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2.Work and Energy Solutions
Gravitational Potential Energy
The work done, W, in lifting a mass against gravity is equal to its gain in potential energy, PE.
m
h
m
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2.Work and Energy Solutions
Kinetic Energy
A mass in motion has an amount of Kinetic Energy proportional to the square of the velocity
F
vi
vf
m
m
d
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2.Work and Energy Solutions
mg
Fperp= mgcosθ
FN = table pushes box up
105cos8= 1.0 x 104N
F Para= mgsinθ accelerating force:
pulls box down ramp
Fperp= mgcosθ
105cos8= 1.0 x 104N
All acceleration down ramp is: a = g sin θ
F Para= mgsinθ 105sin8= 1.4 x 104N
5. The diagram below represents a 155­newton box on a ramp. Applied force F causes the box to slide from point A to point B.
What is the total amount of gravitational potential energy gained by the box? 1. 28.4 J
2. 279 J
3. 868 J
4. 2740 J
As the box moves up the incline, it gains gravitational potential energy.
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2.Work and Energy Solutions
PE / KE Balance for an object descending under the influence of Gravity
PEtop = KE Bottom
PE > KE
Top of the Hill
PE = KE
PE < KE
Halfway down the Hill
Bottom of the Hill
v =0
∆h
Ramp
Hill
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2.Work and Energy Solutions
Assignment: Work & Energy 1. 1. As a box is pushed 30. meters across a horizontal floor by a constant horizontal force of 25 newtons, the kinetic energy of the box increases by 300. Joules. How much total internal energy is produced during this process?
1. 150 J
2. 250 J
3. 450 J
4. 750 J
2. A 15.0­kilogram mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the total work that must be done on the mass to increase its speed to 11.5 meters per second?
1. 120. J
Work was done on the box that 2. 422 J
increased its kinetic energy. The 3. 570. J
difference between the final kinetic 4. 992 J
energy and the initial kinetic energy must be equal to the work done on the box.
3. A child, starting from rest at the top of a playground slide, reaches a speed of 7.0 meters per second at the bottom of the slide. What is the vertical height of the slide? [Neglect friction.]
1. 0.71 m
2. 1.4 m
The potential energy at the top of the slide is converted 3. 2.5 m
into kinetic energy at the 4. 3.5 m
bottom of the slide.
PEtop = KE bottom
4. The total work done in lifting a typical high school physics textbook a vertical distance of 0.10 meter is approximately
1. 0.15 J
Work is done in lifting the textbook 2. 1.5 J
against gravity. Estimate the weight of 3. 15 J
the textbook as 15 N.
4. 150 J
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2.Work and Energy Solutions
6. A 75­kilogram bicyclist coasts down a hill at a constant speed of 12 meters per second. What is the kinetic energy of the bicyclist?
1. 4.5 × 102 J
2. 9.0 × 102 J
3. 5.4 × 103 J
4. 1.1 × 104 J
2
y = kx
7. Which is an SI unit for work done on an object? W = Fd = 1/2m∆v2 = mg∆h Nm or Joules, J, kg*m2/s2
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2.Work and Energy Solutions
Base your answer to this question on the information below.
A boy pushes his wagon at constant speed along a level sidewalk. The graph below represents the relationship between the horizontal force exerted by the boy and the distance the wagon moves.
The area under the F vs d graph is the work done.
8. As the boy pushes the wagon, what happens to the wagon’s energy?
1. Gravitational potential energy increases. Work is being done on the wagon.
2. Gravitational potential energy decreases. The wagon does not change vertical position, so there is no change in its 3. Internal energy increases.
gravitational potential energy. Since 4. Internal energy decreases.
work is being done on the wagon, the internal energy increases.
9. What is the total work done by the boy in pushing the wagon 4.0 meters?
1. 5.0 J
2. 7.5 J
The force is consistent at 30.0 N.
3. 120 J
Reason 4. 180 J
W = Fd = (30.0 N)(4.0 m) = 120 J
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2.Work and Energy Solutions
10. The work done in lifting an apple one meter near Earth’s surface is approximately:
1. 1 J = 1 Nm The work done is in overcoming gravity acting on the apple.
2. 0.01 J
One Joule is the work done when one newton acts 3. 100 J
through a distance of one meter: W = F×d. The apple’s 4. 1000 J
weight is approximately one newton.
*Note: 1kg = 10N
11. PE= max
As a ball falls freely toward the ground, its total mechanical energy:
1. decreases
Think Conservation of Energy!
2. increases
There is no loss or gain in mechanical energy, 3. remains the same
only a transformation of gravitational KE = 0, v =0
potential energy to kinetic energy as the ball falls toward the ground.
12. PE= h = 0
KE = max
A child does 0.20 joule of work to compress the spring in a pop­up toy. If the mass of the toy is 0.010 kilogram, what is the maximum vertical height that the toy can reach after the spring is released?
1. 20. m
The work done on the toy has changed its potential energy.
2. 2.0 m
The work done on the toy is equivalent to the change in its potential energy.
3. 0.20 m
4. 0.020 m
13. Which situation describes a system with decreasing gravitational potential energy?
How does vertical distance related to gravitational 1. a girl stretching a horizontal spring
potential energy?
2. a bicyclist riding up a steep hill
ΔPE = mgΔh. As height increases, so does 3. a rocket rising vertically from Earth
gravitational potential energy. In choice (4), the 4. a boy jumping down from a tree limb height decreases, so the gravitational potential energy also decreases.
14. How much work is required to lift a 10.­newton weight from 4.0 meters to 40. meters above the surface of Earth?
1. 2.5 J
2. 3.6 J
3. 3.6 × 102 J
4. 4.0 × 102 J
15. If the speed of a moving object is doubled, the kinetic energy of the object is
1. halved
If the velocity is doubled, the square of that 2. doubled
velocity would result in an increase in the 3. unchanged
kinetic energy that is 4 times greater (quadrupled).
4. quadrupled
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2.Work and Energy Solutions
16. Base your answer to this question on the information below.
A 65­kilogram pole vaulter wishes to vault to a height of 5.5 meters.
Calculate the minimum amount of kinetic energy the vaulter needs to reach this height if air friction is neglected and all the vaulting energy is derived from kinetic energy. 1.
2.
3.
4.
KE = 3.5 × 103 J
KE = 3.0 × 103 J
KE = 2.5 × 103 J
KE = 4.0 × 103 J
17. What is the maximum amount of work that a 6000.­watt motor can do in 10. seconds? 1. 6.0 × 101 J
2. 6.0 × 102 J
3. 6.0 × 103 J
4. 6.0 × 104 J
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2.Work and Energy Solutions
m = 10 kg
A
PE > KE
Top of the Hill
va = 0
HA=5
C
Hc=2.5
PE = KE
Halfway down the Hill
HB=0
PE < KE
B
Bottom of the Hill
1. Find PEA and KEA: PEA = mgh = 10(9.8)5 = 490 J
KEA = 1/2mv2= 0 J (at rest)
2. Find PEB and KEB: (neglect friction) PEB = mgh = 10(9.8)(0) = 0 J
KEB = PEA = 490 J
Due to conservation of energy
3. Find PEc and KEc: PEC = mgh = 10(9.8)(2.5) = 245 J
KEC = PEC = 245J (halfway down) 4. Find vA:
Given as v = 0 5. Find vB: (Neglect Friction)
6. Find KEB when there is a 20 Joule loss due to friction:
KEB­ Q = 490J ­ 20J = 470J 7. Find vB: (with Friction losses)
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2.Work and Energy Solutions
As the car travels up the hill at constant speed, the kinetic energy remains the same. However, the potential energy increases as it moves up the hill at a higher distance above point A. const. v = const. KE
increasing v = increasing KE
increasing h = increasing PE
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2.Work and Energy Solutions
19. A car with mass m possesses momentum of magnitude p. Which expression correctly represents the kinetic energy, KE, of the car in terms of m and p?
20. A block weighing 40. newtons is released from rest on an incline 8.0 meters above the horizontal, as shown in the diagram below.
If 50. joules of heat is generated as the block slides down the incline, the maximum kinetic energy of the block at the bottom of the incline is 1. 50. J
The potential energy of the block 2. 270 J
at rest minus the heat energy “lost” equals the maximum 3. 320 J
kinetic energy that the block will 4. 3100 J
have at the bottom of the incline.
Weight = 40N (Given) mass = 40N / 9.8 m/s2 = 4 kg
Energy at Top = mgh = 40N * 8m = 320 J
Energy at Bottom (frictionless) = full conversion
Frictionless Speed at Bottom
Speed at Bottom with Friction
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2.Work and Energy Solutions
21. The work done on a slingshot is 40.0 joules to pull back a 0.10­kilogram stone. If the slingshot projects the stone straight up in the air, what is the maximum height to which the stone will rise? [Neglect friction.] 1. 0.41 m
The work done on the stone 2. 41 m
changes its height which changes its potential energy:
3. 410 m
4. 4.1 m
22. While riding a chairlift, a 55­kilogram skier is raised a vertical distance of 370 meters. What is the total change in the skier’s gravitational potential energy? 1. 5.4 × 101 J
2. 5.4 × 102 J
3. 2.0 × 104 J
4. 2.0 × 105 J
23. Student A lifts a 50.­newton box from the floor to a height of 0.40 meter in 2.0 seconds. Student B lifts a 40.­newton box from the floor to a height of 0.50 meter in 1.0 second. Compared to student A, student B does 1. the same work but develops more power
2. the same work but develops less power
3. more work but develops less power
4. less work but develops more power
Apply the equations for work and power to both students. Both students did the same amount of work, 20 J. Student B did that work in half the time as student A. Applying the equation for power, we see that student B expended more power:
24. A book sliding across a horizontal tabletop slows until it comes to rest. Describe what change, if any, occurs in the book’s kinetic energy and internal energy as it slows.
1. The kinetic energy increases and the internal energy decreases.
2. The kinetic energy decreases and the internal energy increases.
3. The kinetic energy decreases and the internal energy decreases.
4. The kinetic energy increases and the internal energy increases.
We know friction can have at least two effects on moving objects. Friction will cause a moving object to slow down or decrease in kinetic energy while also causing the object to heat up which increases the internal energy.
25. Which quantity is a measure of the rate at which work is done?
1. energy
2. power
3. momentum
4. velocity
26. A 110­kilogram bodybuilder, and his 55­kilogram friend, run up identical flights of stairs. The bodybuilder reaches the top in 4.0 seconds while his friend takes 2.0 seconds. Compared to the power developed by the bodybuilder while running up the stairs, the power developed by his friend is
1. the same
2. twice as much
3. half as much
4. four times as much
27. The work done in accelerating an object along a frictionless horizontal surface is equal to the change in the object’s
W = Fd = ΔEt
1. momentum
2. velocity
Work done on an object transfers 3. potential energy
energy to the object. Here, the work 4. kinetic energy
produces an increase in the object's velocity since it is being accelerated. This in turn increases its kinetic energy since KE=1/2mv2
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2.Work and Energy Solutions
28. A box is pushed to the right with a varying horizontal force. The graph below represents the relationship between the applied force and the distance the box moves.
What is the total work done in moving the box 6.0 meters?
1. 9.0 J
2. 18 J
3. 27 J
4. 36 J
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2.Work and Energy Solutions
29. In the graph below, line 2 represents the relationship between vertical height and gravitational potential energy for an object near Earth’s surface.
Which of the lines on the graph represents the relationship between gravitational potential energy and vertical height for an object having a greater mass than the object represented by line 2?
1. Line 1 2. Line 2 3. Line 3 Line 1, with the steeper slope, indicates a greater mass.
30. Base your answer to the question on the graph below, which represents the relationship between vertical height and gravitational potential energy for an object near Earth’s surface. The slope of the graph represents the weight, w = mg, of the object.
What physical quantity does the slope of the graph represent?
1. the weight of the object 2. the mass of the object 3. the change in potential energy of the object 4. the change in height of the object 16
2.Work and Energy Solutions
31. Base your answer to the question on the information and diagram.
A 250.­kilogram car is initially at rest at point A on a roller coaster track. The car carries a 75­kilogram passenger and is 20. meters above the ground at point A. [Neglect friction.]
h = 20
h = 10
h = 0
Compare the total mechanical energy of the car and passenger at points A, B, and C.
1. The total mechanical energy is less at point C than it is at points A or B. 2. The total mechanical energy is greatest at point A. 3. The total mechanical energy is the same at all three points. 4. The total mechanical energy is greatest at point B.
Think conservation of energy. The total mechanical energy will remain the same at all three points, but the amounts of potential and kinetic energies that make up the total mechanical energy will vary. At point A, the potential energy will be greater that than the kinetic energy, while at point B, the kinetic energy will be greater than the potential energy.
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2.Work and Energy Solutions
32. A 1.0­kilogram book resting on the ground is moved 1.0 meter at various angles relative to the horizontal. In which direction does the 1.0­meter displacement produce the greatest increase in the book’s gravitational potential energy?
Potential energy increases as you work against gravity.
The change in gravitational potential energy is ΔPE = mgΔh. The vertical displacement, choice (4), would displace the book by the greatest increase in h, the height above the horizontal.
33. A student does 60. joules of work pushing a 3.0­kilogram box up the full length of a ramp that is 5.0 meters long. What is the magnitude of the force applied to the box to do this work?
1. 20. N 2. 15 N 3. 12 N 4. 4.0 N
34. A 0.50­kilogram ball is thrown vertically upward with an initial kinetic energy of 25 joules. Approximately how high will the ball rise? [Neglect air resistance.]
1. 2.6 m 2. 5.1 m 3. 13 m 4. 25 m
As the ball loses kinetic energy, it gains potential energy.
Apply the law of conservation of energy. We know that the total mechanical energy of the ball is constant. As the ball rises, the loss of kinetic energy as its velocity decreases is matched with a corresponding increase it its gravitational potential energy as its height increases. We have been given the ball's maximum kinetic energy, which must be equal to its maximum gain in potential energy. From the mass of the ball and its maximum kinetic energy, we can calculate the maximum height that it can reach:
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2.Work and Energy Solutions
The power used is related to the work done in a period of time.
The power developed is the work per unit time. Remember that work is force times distance. 19
2.Work and Energy Solutions
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