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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 196: Lesson Check

Exercise 8

Step 1

1 of 2

In this problem, we are asked to state how work is calculated when the force and displacement are in the same direction.

Step 2

2 of 2

When $vec{mathbf{F}}$ and $vec{mathbf{d}}$ are parallel, the work done is simply the product of their magnitudes, that is
$$
boxed{W = Fd}
$$

Exercise 9

Step 1

1 of 2

In this problem, we are asked to identify which component of the force is used to calculate the work done when the force and the displacement are at an angle to each other.

Step 2

2 of 2

The component of the force is $F cos theta$, where $theta$ is the angle between the force and displacement. This component is the component **parallel** to the displacement.

Exercise 10

Step 1

1 of 2

In this problem, we are asked if it is possible to do work on an object that remain at rest.

Step 2

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It is **not possible**. The work done is proportional to the *displacement*, and an object that remains at rest has zero displacement. There is zero work done on the object.

Exercise 11

Step 1

1 of 5

In this problem, we give examples in which friction does (a) positive work, and (b) negative work.

Step 2

2 of 5

### Part A.
The only force that can do positive friction is *static* friction. One example is two blocks on top of each other, and the lower block is pulled with a force in which the upper block does not slip. The static friction would push the upper block in a direction parallel to the applied force and the displacement of the two blocks, so the work done is positive.

Step 3

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Another example of static friction doing positive work is a child riding a sled. The static friction of the sled on the child keeps the child moving forward. The static friction of the sled does positive work on the child.

Step 4

4 of 5

### Part B.
*Kinetic* friction always does negative work. One example is a car that turn off its engine would slow down since the kinetic friction would apply a force opposite the direction of the car’s velocity and displacement, doing negative work.

Step 5

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Another example would be a person sliding across a rough surface. The kinetic friction would always be opposite the displacement so the work done is negative.

Exercise 12

Step 1

1 of 3

In this problem, we are given that a child in a tree house lifts a dog of weight $w = 22~mathrm{N}$ upward for $d = 4.7~mathrm{m}$. We calculate the work done by the child.

Step 2

2 of 3

The force exerted by the child must be equal to the weight of the dog, so $F = w$. We have

$$
begin{align*}
W &= Fd \
&= wd \
&= left( 22~mathrm{N} right) left( 4.7~mathrm{m} right) \
&= 103.4~mathrm{J} \
W &= boxed{ 1.0 times 10^{2}~mathrm{J} }
end{align*}
$$

Result

3 of 3

$$
W = 1.0 times 10^{2}~mathrm{J}
$$

Exercise 13

Step 1

1 of 3

In this problem, a suitcase is moved by a student over a distance $d = 0.95~mathrm{m}$ and the work done is $W = 32~mathrm{J}$. We calculate the force applied by the student.

Step 2

2 of 3

The force and displacement are parallel. From the definition of work, we have

$$
begin{align*}
W &= Fd \
implies F &= frac{W}{d} \
&= frac{32~mathrm{J}}{0.95~mathrm{m}} \
&= 33.68421~mathrm{N} \
F &= boxed{ 34~mathrm{N} }
end{align*}
$$

Result

3 of 3

$$
F = 34~mathrm{N}
$$

Exercise 14

Step 1

1 of 3

In this problem, a farmhand pushes a bale of hay over a displacement of $d = 3.9~mathrm{m}$ horizontally. She applies a force $F = 88~mathrm{N}$ at an angle $theta = 25^{circ}$ below the horizontal. We calculate the work done by the farmhand on the hay.

Step 2

2 of 3

The angle between the force and displacement is $theta$. The work done must be

$$
begin{align*}
W &= vec{mathbf{F}} cdot vec{mathbf{d}} \
&= Fd cos theta \
&= left( 88~mathrm{N} right) left( 3.9~mathrm{m} right) cos 25^{circ} \
&= 311.04483~mathrm{J} \
W &= boxed{ 310~mathrm{J} }
end{align*}
$$

Result

3 of 3

$$
W = 310~mathrm{J}
$$

Exercise 15

Step 1

1 of 4

In this problem, the coefficient of kinetic friction between a large box and the floor is $mu_{k} = 0.21$. If a person pushes it with force $F_{1} = 160~mathrm{N}$ for a distance of $d = 2.3~mathrm{m}$, and the box has mass $m = 72~mathrm{kg}$. We calculate the total work done on the box. We use $g = 9.81~mathrm{m/s^{2}}$.

Step 2

2 of 4

First, we calculate the net force on the box. The frictional force is opposite the applied force, so the net force is

$$
begin{align*}
F &= F_{1} – f \
&= F_{1} – mu_{k}N \
F &= F_{1} – mu_{k}mg
end{align*}
$$

Step 3

3 of 4

From the definition of work, we have

$$
begin{align*}
W &= vec{mathbf{F}} cdot vec{mathbf{d}} \
&= Fd \
&= left( F_{1} – mu_{k}mg right)d \
&= left[ 160~mathrm{N} – left( 0.21 right) left( 72~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right) right] left( 2.3~mathrm{m} right) \
&= 26.84744~mathrm{J} \
W &= boxed{ 27~mathrm{J} }
end{align*}
$$

Result

4 of 4

$$
W = 27~mathrm{J}
$$

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