1. Section 10.6 Using the Law of Conservation of Energy
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2. Using the Law of Conservation of Energy
We can use the law of conservation of energy to develop a before-and-after perspective for energy conservation:
∆ E =W, i.e. Ef = Ei + W,
Kf + (Ug)f + (Us)f + ∆Eth = Ki + (Ug)i + (Us)i + W
This is analogous to the before-and-after approach used with conservation of momentum.
In an isolated system, W = 0:
Kf + (Ug)f + (Us)f + ∆Eth = Ki + (Ug)i + (Us)i
Note: We always want the change ∆Eth in our equations, since we have no way to calculate (Eth )i or (Eth)f .
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3. Choosing an Isolated System
Text: p. 300
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4. QuickCheck 10.18
A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
Answer: D
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5. QuickCheck 10.18
A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
Conservation of energy:
Double x  double v
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6. QuickCheck 10.19
A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball’s launch speed will be
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
Answer: C
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7. QuickCheck 10.19
A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball’s launch speed will be
1.0 m/s
2.0 m/s
2.8 m/s
4.0 m/s
16.0 m/s
Conservation of energy:
Double k  increase v by square root of 2
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8. Example 10.11 Speed at the bottom of a water slide
While at the county fair, Katie tries the water slide. The starting point is 9.0 m above the ground. She pushes off with an initial speed of 2.0 m/s. If the slide is frictionless, how fast will Katie be traveling at the bottom?
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9. Example 10.11 Speed at the bottom of a water slide
While at the county fair, Katie tries the water slide, whose shape is shown in the figure. The starting point is 9.0 m above the ground. She pushes off with an initial speed of 2.0 m/s. If the slide is frictionless, how fast will Katie be traveling at the bottom?
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10. Example 10.11 Speed at the bottom of a water slide (cont.)
prepare Table 10.2 showed that the system consisting of Katie and the earth is isolated because the normal force of the slide is perpendicular to Katie’s motion and does no work. If we assume the slide is frictionless, we can use the conservation of mechanical energy equation. solve Conservation of mechanical energy gives Kf + (Ug)f = Ki + (Ug)i or
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11. Example 10.11 Speed at the bottom of a water slide (cont.)
Taking yf = 0 m, we have which we can solve to get Notice that the shape of the slide does not matter because gravitational potential energy depends only on the height above a reference level.
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12. Example 10.13 Pulling a bike trailer
Monica pulls her daughter Jessie in a bike trailer. The trailer and Jessie together have a mass of 25 kg. Monica starts up a 100-m-long slope that’s 4.0 m high. On the slope, Monica’s bike pulls on the trailer with a constant force of 8.0 N. They start out at the bottom of the slope with a speed of 5.3 m/s. What is their speed at the top of the slope?
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13. Example 10.13 Pulling a bike trailer (cont.)
prepare Taking Jessie and the trailer as the system, we see that Monica’s bike is applying a force to the system as it moves through a displacement; that is, Monica’s bike is doing work on the system. Thus we’ll need to use the full version of Equation , including the work term W.
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14. Example 10.13 Pulling a bike trailer (cont.)
solve If we assume there’s no friction, so that ΔEth = 0, then Equation is Kf + (Ug)f = Ki + (Ug)i + W or
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15. Example 10.13 Pulling a bike trailer (cont.)
Taking yi = 0 m and writing W = Fd, we can solve for the final speed: from which we find that vf = 3.7 m/s. Note that we took the work to be a positive quantity because the force is in the same direction as the displacement. assess A speed of 3.7 m/s—about 8 mph—seems reasonable for a bicycle’s speed. Jessie’s final speed is less than her initial speed, indicating that the uphill force of Monica’s bike on the trailer is less than the downhill component of gravity.
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16. Energy and Its Conservation
Text: p. 304
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17. Section 10.7 Energy in Collisions
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18. Energy in Collisions
A collision in which the colliding objects stick together and then move with a common final velocity is a perfectly inelastic collision.
A collision in which mechanical energy is conserved is called a (perfectly) elastic collision.
While momentum is conserved in all collisions, mechanical energy is only conserved in a perfectly elastic collision.
In an inelastic collision, some mechanical energy is converted to thermal energy.
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19. Example 10.15 Energy transformations in a perfectly inelastic collision
The figure shows two train cars that move toward each other, collide, and couple together. In Example 9.8, we used conservation of momentum to find the final velocity shown in the figure from the given initial velocities. How much thermal energy is created in this collision?
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20. Example 10.15 Energy transformations in a perfectly inelastic collision (cont.)
prepare We’ll choose our system to be the two cars. Because the track is horizontal, there is no change in potential energy. Thus the law of conservation of energy, Equation 10.18a, is Kf + ΔEth = Ki. The total energy before the collision must equal the total energy afterward, but the mechanical energies need not be equal.
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21. Example 10.15 Energy transformations in a perfectly inelastic collision (cont.)
solve The initial kinetic energy is Because the cars stick together and move as a single object with mass m1 + m2, the final kinetic energy is
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22. Example 10.15 Energy transformations in a perfectly inelastic collision (cont.)
From the conservation of energy equation on the previous slide, we find that the thermal energy increases by ΔEth = Ki  Kf = 4.7  104 J  1900 J = 4.5  104 J This amount of the initial kinetic energy is transformed into thermal energy during the impact of the collision. assess About 96% of the initial kinetic energy is transformed into thermal energy. This is typical of many real-world collisions.
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23. Elastic Collisions (simple example)
Elastic collisions obey conservation of momentum and conservation of mechanical energy. Consider the simple case where one object starts at rest.
Momentum:
Energy:
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24. Example 10.16 Velocities in an air hockey collision
On an air hockey table, a moving puck, traveling to the right at 2.3 m/s, makes a head-on collision with an identical puck at rest. What is the final velocity of each puck, assuming this collision was approximately elastic?
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25. Example 10.16 Velocities in an air hockey collision (cont.)
prepare The before-and-after visual overview is shown in the figure. We’ve shown the final velocities in the picture, but we don’t really know yet which way the pucks will move. Because one puck was initially at rest, we can use Equations to find the final velocities of the pucks. The pucks are identical, so we have m1 = m2 = m.
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26. Example 10.16 Velocities in an air hockey collision (cont.)
solve We use Equations with m1 = m2 = m to get The incoming puck stops dead, and the initially stationary puck goes off with the same velocity that the incoming one had.
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27. Example 10.17 Protecting your head
A bike helmet—basically a shell of hard, crushable foam—is tested by being strapped onto a 5.0 kg dummy head and dropped from a height of 2.0 m onto a hard anvil. What force is encountered by the dummy head if the impact crushes the foam by 3.0 cm?
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28. Example 10.17 Protecting your head
prepare We can use the work-energy equation to calculate the force on the dummy head. We’ll choose the dummy head and the earth to be the system; the foam in the helmet is part of the environment. We make this choice so that the force on the dummy head due to the foam is an external force that does work W on the dummy head.
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29. Example 10.17 Protecting your head (cont.)
solve The head starts at rest, speeds up as it falls, then returns to rest during the impact. Overall, then, Kf = Ki. Furthermore, ΔEth = 0 because there’s no friction to increase the thermal energy. Only the gravitational potential energy changes, so the work-energy equation is (Ug)f  (Ug)i = W The upward force of the foam on the head is opposite the downward displacement of the head. Referring to Tactics Box 10.1, we see that the work done is negative: W = Fd, where we’ve assumed that the force is constant. Using this result in the work-energy equation and solving for F, we find
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30. Example 10.17 Protecting your head (cont.)
Taking our reference height to be y = 0 m at the anvil, we have (Ug)f = 0. We’re left with (Ug)i = mgyi, so This is the force that acts on the head to bring it to a halt in 3.0 cm. More important from the perspective of possible brain injury is the head’s acceleration: assess The accepted threshold for serious brain injury is around 300g, so this helmet would protect the rider in all but the most serious accidents. Without the helmet, the rider’s head would come to a stop in a much shorter distance and thus be subjected to a much larger acceleration.
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31. Section 10.8 Power
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32. Power Power is the rate at which energy is transformed or transferred.
The unit of power is the watt:
1 watt = 1 W = 1 J/s
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33. Output Power of a Force A force doing work transfers energy.
The rate that this force transfers energy is the output power of that force:
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34. Example 10.18 Power to pass a truck
Your 1500 kg car is behind a truck traveling at 60 mph (27 m/s). To pass the truck, you speed up to 75 mph (34 m/s) in 6.0 s. What engine power is required to do this? prepare Your engine is transforming the chemical energy of its fuel into the kinetic energy of the car. We can calculate the rate of transformation by finding the change ΔK in the kinetic energy and using the known time interval.
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35. Example 10.18 Power to pass a truck (cont.)
solve We have so that ΔK = Kf  Ki = (8.67  105 J)  (5.47  105 J) = 3.20  105 J To transform this amount of energy in 6 s, the power required is This is about 71 hp. This power is in addition to the power needed to overcome drag and friction and cruise at 60 mph, so the total power required from the engine will be even greater than this.
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36. QuickCheck 10.20
Four students run up the stairs in the time shown. Which student has the largest power output?
Answer: B
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37. QuickCheck 10.20
Four students run up the stairs in the time shown. Which student has the largest power output? B.
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38. QuickCheck 10.21
Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power?
Answer: A
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39. QuickCheck 10.21
Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power?
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40. Bonus Material: Angular Momentum of a Particle
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41. Angular Motion of a Particle Moving in a Straight Line
A particle moving in a straight line sweeps out an angle relative to a fixed point (the pivot or axis of rotation), and hence has angular velocity, angular acceleration, etc.
For r the distance to the pivot, 𝜙 is the angle between the radial line and the linear velocity (with magnitude v), or linear acceleration (magnitude a), the angular speed and angular acceleration are given by
ωr = v sin 𝜙 αr = a sin 𝜙
The angular momentum is given by (where m is the mass of the particle)
L = m v r sin 𝜙
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42. Example: Boy and a Merry-go-round
A 35 kg boy runs due north with a speed of 4.0 m/s, and jumps onto a merry-go-round which is initially at rest, landing at a point 1.2 m directly to the right of the center of the merry-go-round. The merry-go-round has a moment of inertia of 550 kg m2. What is the angular speed of the merry-go-round after the boy has jumped on?
0.29 rad/s
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