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College Physics by Openstax Chapter 7 Problem 6

How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in Figure 7.33? Assume no friction acts on the wagon.

Figure 7.33 The boy does work on the system of the wagon and the child when he pulls them as shown.

Solution:

The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle \theta between them. In symbols,

W=Fd \cos \theta

In this case, we are given the following values:

\begin{align*}
F & = 50\ \text{N} \\
d & = 30\ \text{m} \\
\theta & = 30^{\circ} 
\end{align*}

Substituting these values into the equation, we have

\begin{align*}
W & = Fd \cos \theta \\
W & = \left( 50\ \text{N} \right)\left( 30\ \text{m} \right) \cos 30^{\circ } \\
W & = 1299.0381\ \text{N} \cdot \text{m} \\
W & = 1.30 \times 10^{3}\ \text{N} \cdot \text{m} \\
W & = 1.30 \times 10^{3}\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

College Physics by Openstax Chapter 7 Problem 5

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º with the horizontal. (See Figure 7.32.) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.

Figure 7.32 A man pushes a crate up a ramp.

Solution:

The Work Done by the Man on the Crate

The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle \theta between them. In symbols,

W=Fd \cos \theta

In case where the work done by the man to the crate, the following values are given:

\begin{align*}
F = & \ 500\ \text{N} \\
d = & \ 4\ \text{m} \\
\theta = & \ 0^{\circ} \color{Blue} \left( \text{Force is parallel to displacement} \right) 
\end{align*}

Substituting these values in the equation, we have

\begin{align*}
W = & \ Fd \cos \theta \\
W = & \ \left( 500\ \text{N} \right) \left( 4\ \text{m} \right) \cos 0^{\circ} \\
W = & \ 2000\ \text{N} \cdot \text{m}
\end{align*}

The work done by the man on his body

In this case, the force exerted is counteracted by the weight of the man. This force is directed upward. The displacement is still the 4.0 m along the inclined plane. The angle between the force and the displacement is 70 degrees. 

\begin{align*}
W = & \ Fd \cos \theta \\
W = & \ mg d \cos \theta \\
W = & \ \left( 85.0\ \text{kg} \right) \left( 9.80\ \text{m/s}^2 \right)\left( 4.0\ \text{m} \right) \cos 70^{\circ} \\
W = & \ 1139.6111\ \text{N} \cdot \text{m}
\end{align*}

The Total Work

The total work done by the man is the sum of the work he did on the crate and on his body.

\begin{align*}
W_{T} & = 2000\ \text{N}\cdot \text{m} + 1139.6111\ \text{N}\cdot \text{m} \\
W_{T} & = 3139.6111 \ \text{N}\cdot \text{m} \\
W_{T} & = 3.14 \times 10^{3} \ \text{N}\cdot \text{m} \\
W_{T} & = 3.14 \times 10^{3} \ \text{J}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

College Physics by Openstax Chapter 7 Problem 4

Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See Table 7.1 for the energy content of gasoline.) (a) What is the magnitude of the force exerted to keep the car moving at constant speed? (b) If the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s?

Solution:

Part A

According to Table 7.1, the energy in 1 gallon of gasoline is 1.2 \times 10^{8}\ \text{J}. Since only 30% of the gasoline goes into useful work, the work done by the friction W_{f} is 

\begin{align*}
W_{f} & =0.30 \left( 2.0\ \text{gal} \right)\left( 1.2 \times 10^{8} \ \text{J/gal}\right) \\
W_{f} & = 72 \times 10^{6}\ \text{J}
\end{align*}

Now, the work done by the friction can also be calculated using the formula below, where F_{f} is the magnitude of the friction force that keeps the car moving at constant speed, and d is the distance traveled by the car.

\begin{align*}
W_{f}=F_{f}d
\end{align*}

We can solve for F_{f} in terms of the other variables.

F_{f} = \frac{W_{f}}{d}

Substituting the given values, we can now solve for the unknown magnitude of the force exerted to keep the car moving at constant speed.

\begin{align*}
F_{f} & = \frac{W_{f}}{d} \\
F_{f} & = \frac{72 \times 10^{6}\ \text{J}}{108\ \text{km}} \\
F_{f} & = \frac{72 \times 10^{6}\ \text{N}\cdot \text{m}}{108 \times 10^{3}\ \text{m}} \\
F_{f} & = 666.6667\ \text{N} \\
F_{f} & = 6.7 \times 10^{2}\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

If the required force is directly proportional to speed, then there must be a linear relationship between the required force and speed. In this situation, we can just simply used ratio and proportion to compute for the number of gallons.

\begin{align*}
\frac{2.0\ \text{gal}}{30.0\ \text{m/s}} & = \frac{x}{28.0\ \text{m/s}} \\
x & = \frac{\left( 2.0\ \text{gal} \right)\left( 28.0\ \text{m/s} \right)}{30.0\ \text{m/s}} \\
x & = 1.8667\ \text{gal} \\
x & = 1.9\ \text{gal} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

College Physics by Openstax Chapter 7 Problem 3

(a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. (b) What is the work done on the elevator car by the gravitational force in this process? (c) What is the total work done on the elevator car?

Solution:

The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle \theta between them. In symbols,

W=Fd \cos \theta

Part A

The force in the cable is equal to the combined effect of the weight of the elevator and the friction that opposes the motion. That is

\begin{align*}
F & = mg + f \\
F & = \left( 1500\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right)+100\ \text{N} \\
F & = 14800\ \text{N}
\end{align*}

This force in the cable is directed upward. The displacement is also upward, making the angle between the two quantities equal to zero. Thus, \theta = 0.

Substituting these values in the equation, the work done by the cable is

\begin{align*}
W & = Fd \cos \theta \\
W & = \left( 14\ 800\ \text{N} \right)\left( 40.0\ \text{m} \right) \cos 0^\circ \\
W & = 592\ 000\ \text{J} \\
W & = 5.92 \times 10^{5} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The force due to gravity is equal to the weight of the elevator alone. That is 

\begin{align*}
\text{Weight} & = mg \\
 & = \left( 1\ 500\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right) \\
 & = 14\ 700\ \text{N}
\end{align*}

This force is directed downward, whereas the displacement is directed upward. Therefore, the angle \theta between the two quantities is \theta = 180^\circ.

Substituting these values in the formula for work, we have

\begin{align*}
W & = Fd \cos \theta \\
W & = \left( 14\ 700\ \text{N} \right)\left( 40.0\ \text{m} \right) \cos 180^\circ \\
W & = -588\ 000\ \text{J} \\
W & = -5.88 \times 10^{5}\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

Since the elevator is moving at a constant speed, it is in equilibrium. This means that the net external force experience by the elevator is zero. Therefore, the total work done on the elevator car is 

W_{T} = 0\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer}  \right)

College Physics by Openstax Chapter 7 Problem 2

A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. (Neglect friction in your calculations.)

Solution:

Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy is \Delta PE_{g} = mgh, with h being the increase in height and g the acceleration due to gravity.

W=mgh

We are given the following values: m=75.0\ \text{kg}, g=9.80\ \text{m/s}^2, and h=2.50\ \text{m}.

Substitute the given in the formula.

\begin{align*}
W & = mgh \\
W & = \left( 75.0\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right)\left( 2.50\ \text{m} \right)\\
W & = 1837.5\ \text{Nm} \\
W & = 1837.5\ \text{J} \\
W & = 1.84 \times 10 ^{3} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The work done is about 1.84 \times 10 ^ {3}\ \text{Joules} .

College Physics by Openstax Chapter 7 Problem 1

How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories.

Solution:

The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of the displacement, times the cosine of the angle \theta between them. In symbols,

W=Fd \cos \theta

We are given the following values: F=5.00\ \text{N}, d=0.600\ \text{m}, and \theta=0^\circ.

Substitute the given values in the formula for work.

\begin{align*}
W & = Fd \cos \theta \\
W & = \left( 5.00\ \text{N} \right)\left( 0.600\ \text{m} \right) \cos 0^\circ \\
W & = 3.00\ \text{Nm} \\
W & = 3.00\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The work done is 3.00 Joules. Now, we can convert this in unit of kilocalories knowing that 1\ \text{kcal} = 4186\ \text{J}. 

\begin{align*}
W & = 3.00\ \text{J} \\
W & = 3.00\ \text{J}\ \times \ \frac{1\ \text{kcal}}{4186\ \text{J}} \\
W & = 0.000717\ \text{kcal} \\
W & = 7.17 \times 10 ^{-4} \ \text{kcal} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The work done in kilocalories is about 7.17 \times 10 ^{-4}.

Problem 6-19: The angular velocity of an “artificial gravity”

A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of 9.80 m/s2 at the rim?

Solution:

We are given the following quantities:

\text{radius} = \frac{\text{diameter}}{2} = \frac{200\ \text{m}}{2} = 100 \ \text{m}
\text{centripetal acceleration}, a_c = 9.80 \ \text{m/s}^2

Centripetal acceleration is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. The formula for centripetal acceleration is

a_{c} = r \omega ^2

If we solve for the angular velocity in terms of the other quantities, we have

\omega = \sqrt{\frac{a_c}{r}}

Substituting the given quantities,

\begin{align*}
\omega & = \sqrt{\frac{a_c}{r}}  \\ \\
\omega & = \sqrt{\frac{9.80 \ \text{m/s}^2}{100\ \text{m}}} \\ \\
\omega & = 0.313 \ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 3 Problem 4

Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements \vec{A} and \vec{B} , as in Figure 3.53, then this problem asks you to find their sum \vec{R}=\vec{A}+\vec{B} .)

Figure 3.53

Solution:

Figure 3.4A

Consider Figure 3.54A.

The resultant of the two vectors \vec{A} and \vec{B} is labeled \vec{R}. This \vec{R} is directed \theta ^{\circ} from the x-axis.

We shall use the right triangle formed to solve for the unknowns.

Solve for the magnitude of the resultant.

\begin{align*}
R & = \sqrt{A^2 +B^2} \\
R & = \sqrt{\left(18.0 \ \text{m}  \right)^2+\left( 25.0 \ \text{m} \right)^2} \\
R & = 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Solve for the value of \theta .

\begin{align*}
\theta & = \arctan \left( \frac{B}{A} \right) \\
\theta & = \arctan \left( \frac{25.0 \ \text{m}}{18.0 \ \text{m}} \right) \\
\theta & = 54.2^\circ 
\end{align*}

We need the complementary angle for the compass angle. 

\begin{align*}
90^\circ -54.2^\circ =35.8^\circ 
\end{align*}

Therefore, the compass angle reading is

\begin{align*}
35.8^\circ , \text{W of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
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College Physics by Openstax Chapter 3 Problem 2

Find the following for path B in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side

Solution:

Part A

The total distance traveled is 

\begin{align*}

\text{d} & = \left(4 \times 120 \text{m} \right) + \left(3 \times 120\ \text{m} \right) + \left(3 \times 120\ \text{m} \right) \\
\text{d} & = 1 200\ \text{m}  \ \qquad \  {\color{DarkOrange} \left( \text{Answer} \right)}\\

\end{align*}

Part B

The magnitude of the displacement is 

\begin{align*}
\text{s} & = \sqrt{\left( s_x \right)^2+\left( s_y \right)^2} \\
\text{s} & = \sqrt{\left( 1 \times 120\ \text{m} \right)^2+ \left( 3 \times 120 \ \text{m} \right)^2} \\
\text{s} & = \sqrt{\left( 120\ \text{m} \right)^2+ \left( 360 \ \text{m} \right)^2} \\
\text{s} & = 379 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

The direction is

\begin{align*}
\theta & = \arctan \left( \frac{s_y}{s_x} \right) \\
\theta & = \arctan \left( \frac{360\ \text{m}}{120\ \text{m}} \right) \\
\theta & = 71.6^\circ , \ \text{N of E} \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}
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College Physics 2.51 – Time of the hiker to move out from a falling rock

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution:

Part A

We know that the initial height, y_0 of the rock is 105 meters, and the initial velocity, v_0 is zero. We shall solve for the distance traveled by the rock for 1.5 seconds from the initial position first to find the height at detection.

The change in height is

\displaystyle \begin{aligned}
\Delta \text{y}&=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2 \\
&=\left( 0  \right)\left( 1.50 \ \text{s} \right)+\frac{1}{2}\left( 9.81\ \text{m/s}^{2} \right)\left( 1.50\ \text{s} \right)^{2}\\
&=0+11.036\ \text{m} \\
&=11.04 \ \text{m} 
\end{aligned}

So, the rock falls about 11.04 m from the initial height for 1.50 seconds. Therefore, the height of the rock above his head at this point is

\displaystyle \begin{aligned}
\text{y}&=\text{y}_{0}-\Delta \text{y} \\
&=105\ \text{m}-11.04\ \text{m} \\
&=93.96 \ \text{m}
\end{aligned}

Part B

We shall solve for the total time of travel, that is, from the initial position to his head. Then we shall subtract 1.50 s from that to solve for the unknown time of moving out. The total time of travel is

\begin{aligned}
\text{y} & =\frac{1}{2}\text{at}^{2} \\
&\text {Solving for t, we have}\\
\text{t}&=\sqrt{\frac{\text{2y}}{\text{a}}} \\
&=\sqrt{\frac{2\left( 105\ \text{m} \right)}{9.81 \ \text{m/s}^{2}}} \\
&=4.63 \ \text{s}

\end{aligned}

Therefore, to move out the hiker has about

\begin{aligned}
\text{t}&=4.63 \ \text{s}-1.50\ \text{s}\\
&=3.13\ \text{s}
\end{aligned}