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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 211: Lesson Check

Exercise 40

Step 1

1 of 4

In this problem, the potential energy of an object decreases by $10~mathrm{J}$. We calculate the chage in the kinetic energy, assuming that there is no friction in the system.

Step 2

2 of 4

Since the potential energy decreases, its change mustbe negative.

$$
Delta PE = -10~mathrm{J}
$$

Step 3

3 of 4

We use the equation for conservation of mechanical energy to find $Delta KE$.

$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
KE_text{i} – KE_text{f} &= PE_text{f} – PE_text{i} \
-left( KE_text{f} – KE_text{i} right) &= PE_text{f} – PE_text{i} \
-Delta KE &= Delta PE \
implies Delta KE &= – Delta PE\
&= -left( -10~mathrm{J} right) \
Delta KE &= boxed{ 10~mathrm{J} }
end{aligned}
$$

Result

4 of 4

$$
Delta KE = 10~mathrm{J}
$$

Exercise 41

Step 1

1 of 2

In this problem, we are asked the condition for the conservation of mechanical energy of a system.

Step 2

2 of 2

The mechanical energy of system is conserved if **only conservative forces** act on the system. Nonconservative forces such as friction and air drag would convert some mechanical energy into other forms, hence the mechanical energy is no longer conserved.

Exercise 42

Step 1

1 of 2

In this problem, we discuss the energy conversions of when a person performs a pole vault.

Step 2

2 of 2

Before the pole vault, the person is converting some chemical energy into mechanical energy of the muscles in order to run. During the pole vault, the person converts some more chemical energy in order to bend the pole. When the pole bends, it stores from spring potential energy. As the person rises when the pole returns to its original shape, the stored spring potential energy is converted to kinetic energy to move upwards. As the person rises, the kinetic energy is converted to gravitational potential energy. After the person reaches the peak, they converted the gravitational potential energy into kinetic energy until hitting the compressional pad. Upon landing and squihing the compressional pad, some of the kinetic energy is transformed into internal energy of the person and the pad.

Exercise 43

Step 1

1 of 5

In this problem, a ball is thrown straight up into the air. It reaches a maximum height and returns to the hand that threw it up. We find the locations in which the kinetic energy is (a) a maximum and (b) minimum. We also find the locations in which the potential energy is (c) maximum and (d) minimum.

Step 2

2 of 5

#### Part A.

The kinetic energy is maximum when its speed is maximum. This happens right after it was thrown up from the hand and when it returns back to the hand.

Step 3

3 of 5

#### Part B.

The kinetic energy is minimum when the speed is minimum. The speed is zero at the peak, so the kinetic energy there must be the minimum.

Step 4

4 of 5

#### Part C.

The potential energy is proportional to the height of the ball. The maximum height is at the peak, so the potential energy is maximum when it is on its peak.

Step 5

5 of 5

#### Part D.

The potential energy is minimum at its lowest elevation. The lowest elevation is at the hand of the thrower.

Exercise 44

Step 1

1 of 3

In this problem, a swimmer starts at rest, slides without friction, and descends through a vertical height of $h = 2.31~mathrm{m}$. We calculate the speed at the bottom of the slide. We use $g – 9.81~mathrm{m/s^{2}}$.

Step 2

2 of 3

The initial kinetic energy is zero since the swimmer is at rest, and the final potential energy is zero since the swimmer is on the “zero potential” elevation. The equation of conservation of mechanical energy gives

$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
0 + mgh &= frac{1}{2}mv_text{f}^{2} + 0 \
v_text{f}^{2} &= 2gh \
implies v_text{f} &= sqrt{2gh} \
&= sqrt{2 left( 9.81~mathrm{m/s^{2}} right) left( 2.31~mathrm{m}right)} \
&= 6.73218~mathrm{m/s} \
v_text{f} &= boxed{6.73~mathrm{m/s}}
end{aligned}
$$

Result

3 of 3

$$
v_text{f} = 6.73~mathrm{m/s}
$$

Exercise 45

Solution 1

Solution 2

Step 1

1 of 5

In this problem, an apple of $m = 0.21~mathrm{kg}$ falls froma tree to the ground, from height $h_text{i} = 4.0~mathrm{m}$. We calculate the kinetic energy, potential energy, and total mechanical energy of the system when the apple is at height $h_text{f} = 3.0~mathrm{m}$ above the ground. We use $g = 9.81~mathrm{m/s^{2}}$.

Step 2

2 of 5

We use the equation of conservation of mechanical energy. The initial speed is zero, so the initial kinetic energy must also be zero. We have

$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{F} \
0 + mgh_text{i} &= KE_text{f} + mgh_text{f} \
implies KE_text{f} &= mg left( h_text{i} – h_text{f} right) \
&= left( 0.21~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right) left( 4.0~mathrm{m} – 3.0~mathrm{m} right) \
&= 2.06010~mathrm{J} \
KE_text{f} &= boxed{ 2.1~mathrm{J} }
end{aligned}
$$

Step 3

3 of 5

The potential energy, from the definition, is

$$
begin{aligned}
PE_text{f} &= mgh_text{f} \
&= left( 0.21~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right) left( 3.0~mathrm{m} right) \
&= 6.18030~mathrm{J} \
PE_text{f} &= boxed{ 6.2~mathrm{J} }
end{aligned}
$$

Step 4

4 of 5

The total mechanical energy is the sum of the kinetic and potential energy. We have

$$
begin{aligned}
E &= KE_text{f} + PE_text{f} \
&= 2.06010~mathrm{J} + 6.18030~mathrm{J} \
&= 8.24040~mathrm{J} \
E &= boxed{ 8.2~mathrm{J} }
end{aligned}
$$

Result

5 of 5

$$
begin{aligned}
KE_text{f} &= 2.1~mathrm{J} \
PE_text{f} &= 6.2~mathrm{J} \
E &= 8.2~mathrm{J}
end{aligned}
$$

Step 1

1 of 5

The problem requires us to get the gravitational Potential Energy, Kinetic Energy, and the total Mechanical Energy of the apple at the height of 3.0m above ground.
to get the Potential Energy at 3.0m, we use

$$
PE_{gravity,f} = mgh = (0.21kg)(9.81m/s^2)(3m)
$$

$$
PE_{gravity,f} = 6.18J
$$

Step 2

2 of 5

To get the Kinetic and total Mechanical Energy at 3.0m above ground, we must first compute the gravitational Potential, Kinetic, and total Mechanical Energy of the apple 4.0m above ground where the apple is at rest and started to fall.
At 4.0m above ground, the apple is at rest therefore,

$$
KE_{i} = 0
$$

For the gravitational Potential Energy,

$$
PE_{gravity,i} = mgh = (0.21kg)(9.81m/s^2)(4.0m)
$$

$$
PE_{gravity,i} = 8.24J
$$

Step 3

3 of 5

We can now compute the total Mechanical Energy

$$
ME = PE + KE = 8.24J + 0
$$

$$
ME = 8.24J
$$

Step 4

4 of 5

To solve for the Kinetic Energy at 3.0m above ground, we will use the computed ME and $PE_{gravity,f}$

$$
ME = PE_{gravity,f} + KE_{f}
$$

Rearranging,

$$
KE_{f} = ME – PE_{gravity,f}
$$

$$
KE_{f} = 8.24J – 6.18J
$$

$$
KE_{f} = 2.06J
$$

Result

5 of 5

$$
KE_{f} = 2.06J
$$

$$
PE_{gravity,f} = 6.18J
$$

$$
ME = 8.24J
$$

Exercise 46

Step 1

1 of 4

In this problem, a baseball glove is thrown straight upward. Its initial kinetic energy is $KE$, and reaches a maximum height of $h$. We calculate the kinetic energy when the height is $h/2$.

Step 2

2 of 4

The potential energy at height $h$ is $mgh$. This must be equal to the initial kinetic energy since the glove has zero speed at this height.

$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
KE + 0 &= 0 + mgh \
implies mgh &= KE
end{aligned}
$$

Step 3

3 of 4

When the height is $h/2$, the equation of conservation of energy is

$$
begin{aligned}
KE_text{i} + PE_text{i} &= KE_text{f} + PE_text{f} \
KE + 0 &= KE_text{f} + frac{1}{2}mgh \
KE &= KE_text{f} + frac{1}{2}KE \
implies KE_text{f} &= frac{1}{2}KE
end{aligned}
$$

Result

4 of 4

$$
KE_text{f} = frac{1}{2}KE
$$

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