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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

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:

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:

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 $7.17 \times 10 ^{-4}$ .

College Physics by Openstax Chapter 6 Problem 31

The Speed of the Roller Coaster at the Top of the Loop

Problem:

Modern roller coasters have vertical loops like the one shown in Figure 6.35. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?

**Figure 6.35** Teardrop-shaped loops are used in the latest roller coasters so that the radius of curvature gradually decreases to a minimum at the top. This means that the centripetal acceleration builds from zero to a maximum at the top and gradually decreases again. A circular loop would cause a jolting change in acceleration at entry, a disadvantage discovered long ago in railroad curve design. With a small radius of curvature at the top, the centripetal acceleration can more easily be kept greater than g
so that the passengers do not lose contact with their seats, nor do they need seat belts to keep them in place.

Solution:

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College Physics by Openstax Chapter 6 Problem 30

The Ideal Speed and the Minimum Coefficient of Friction in Icy Mountain Roads

Problem:

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).

(a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º.

(b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?

Solution:

Part A

The formula for an ideally banked curve is

\tan \theta = \frac{v^2}{rg}

Solving for $v$ in terms of all the other variables, we have

v= \sqrt{rg \tan \theta}

For this problem, we are given

radius, $r=100\ \text{m}$
acceleration due to gravity, $g=9.81\ \text{m/s}^2$
banking angle, $\theta = 15.0^\circ$

Substituting all these values in the formula, we have

\begin{align*} v & = \sqrt{rg \tan \theta} \\ v & = \sqrt{\left( 100\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)\tan 15.0^\circ }\\ v & = 16.2129\ \text{m/s} \\ v & = 16.2\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Let us draw the free-body diagram of the car.

Summing forces in the vertical direction, we have

N \cos \theta + f \sin \theta -w= 0 \quad \quad \quad \color{Blue} \text{Equation 1}

Summing forces in the horizontal directions taking to the left as the positive since the centripetal force is directed this way, we have

N \sin \theta - f \cos \theta = F_c \quad \quad \quad \color{Blue} \text{Equation 2}

We are given the following quantities:

radius of curvature, $r=100\ \text{meters}$
banking angle, $\theta = 15.0^\circ$
velocity, $v=20\ \text{km/h} = 5.5556\ \text{m/s}$
We also know that the friction, $f=\mu_{s} N$ and weight, $w=mg$

We now use equation 1 to solve for N in terms of the other variables.

\begin{align*} N \cos \theta + f \sin \theta -w & = 0 \\ N \cos \theta +\mu_s N\sin \theta & = mg \\ N \left( \cos \theta + \mu_s \sin\theta \right) & =mg \\ N & = \frac{mg}{\cos \theta + \mu_s \sin\theta} \ \qquad \ \color{Blue} \left( \text{Equation 3} \right) \end{align*}

We also solve for N in equation 2.

\begin{align*} N \sin \theta - \mu_s N \cos \theta & = m \frac{v^2}{r} \\ N \left( \sin \theta-\mu_s \cos \theta \right) & = m \frac{v^2}{r} \\ N & = \frac{mv^2}{r \left( \sin \theta-\mu_s \cos \theta \right)} \ \qquad \ \color{Blue} \left( \text{Equation 4} \right) \end{align*}

Now, we have two equations of $N$ . We now equate these two equations.

\frac{mg}{\cos \theta + \mu_s \sin\theta} = \frac{mv^2}{r \left( \sin \theta-\mu_s \cos \theta \right)}

We can now use this equation to solve for $\mu_s$ .

\begin{align*} \frac{\bcancel{m}g}{\cos \theta + \mu_s \sin\theta} & = \frac{\bcancel{m}v^2}{r \left( \sin \theta-\mu_s \cos \theta \right)} \\ rg\left( \sin \theta-\mu_s \cos \theta \right) & = v^2 \left( \cos \theta + \mu_s \sin\theta \right) \\ rg \sin \theta - \mu_s rg \cos \theta & = v^2 \cos \theta +\mu_s v^2 \sin \theta \\ \mu_s v^2 \sin \theta + \mu _s rg \cos \theta & = rg \sin \theta - v^2 \cos \theta \\ \mu _s \left( v^2 \sin \theta + rg \cos \theta \right) & = rg \sin \theta - v^2 \cos \theta \\ \mu _s & = \frac{rg \sin \theta - v^2 \cos \theta}{v^2 \sin \theta + rg \cos \theta} \end{align*}

Now that we have an equation for $\mu_s$ , we can substitute the given values.

\begin{align*} \mu _s & = \frac{rg \sin \theta - v^2 \cos \theta}{v^2 \sin \theta + rg \cos \theta} \\ \mu_s & = \frac{100\ \text{m}(9.81\ \text{m/s}^2) \sin 15.0^\circ -\left( 5.5556\ \text{m/s} \right)^2 \cos 15.0^\circ }{\left( 5.5556\ \text{m/s} \right)^2 \sin 15.0^\circ +100\ \text{m}\left( 9.81\ \text{m/s}^2 \right) \cos 15.0^\circ } \\ \mu_s & = 0.2345 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

College Physics by Openstax Chapter 6 Problem 29

The centripetal acceleration of a large centrifuge as experienced in rocket launches and atmospheric reentries of astronauts

Problem:

A large centrifuge, like the one shown in Figure 6.34(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.

(a) At what angular velocity is the centripetal acceleration $10g$ if the rider is 15.0 m from the center of rotation?

(b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 6.34(b). At what angle $\theta$ below the horizontal will the cage hang when the centripetal acceleration is $10g$ ? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle $10g$ should be.)

**Figure 6.34** (a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and reentries. (credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to always be along its axis.

Solution:

Part A

The centripetal acceleration, $a_c$ , is calculated using the formula $a_c = r \omega ^2$ . Solving for the angular velocity, $\omega$ , in terms of the other variables, we should come up with

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

We are given the following values:

centripetal acceleration, $a_c = 10g = 10 \left( 9.81\ \text{m/s}^2 \right) = 98.1\ \text{m/s}^2$

radius of curvature, $r = 15.0\ \text{m}$

Substituting the given values into the equation,

\begin{align*} \omega & = \sqrt{\frac{a_c}{r}} \\ \\ \omega & = \sqrt{\frac{98.1\ \text{m/s}^2}{15.0\ \text{m}}} \\ \\ \omega & = 2.5573\ \text{rad/sec} \\ \\ \omega & = 2.56\ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The free-body diagram of the force is shown

The free-body diagram of the rider’s cage that hangs on a pivot at the end of the arm of a large centrifuge. College Physics Problem 6-29 — The free-body diagram of the rider’s cage hangs on a pivot at the end of the arm of a large centrifuge.

Summing forces in the vertical direction, we have

\begin{align*} \sum_{}^{} F_y & = 0 \\ \\ F_{arm} \sin \theta-w & = 0 \\ \\ F_{arm} & = \frac{w}{\sin \theta} \ \quad \quad \color{Blue} \text{Equation 1} \end{align*}

Now, summing forces in the horizontal direction, taking into account that $F_c$ is the centripetal force which is the net force. That is,

\begin{align*} F_c & = m a_c \end{align*}

We know that $F_c$ is equal to the horizontal component of the force $F_{arm}$ . That is $F_c = F_{arm} \cos \theta$ . Therefore,

\begin{align*} F_{arm} \cos \theta & = m a_c \\ \end{align*}

Now, we can substitute equation 1 into the equation, and the value of the centripetal acceleration given at $10g$ . Also, we note that the weight $w$ is equal to $mg$ . So, we have

\begin{align*} F_{arm} \cos \theta & = m a_c \\ \\ \frac{w}{\sin \theta} \cos \theta & = m (10g) \\ \\ \frac{mg \cos \theta}{\sin \theta} & = 10 mg \\ \\ \end{align*}

From here, we are going to use the trigonometric identity $\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}$ . We can also cancel $m$ , and g since they can be found on both sides of the equation.

\begin{align*} \frac{1}{\tan \theta} & = 10 \\ \\ \tan \theta & = \frac{1}{10} \\ \\ \theta & = \tan ^{-1} \left( \frac{1}{10} \right) \\ \\ \theta & = 5.71 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

College Physics by Openstax Chapter 6 Problem 27

The radius and centripetal acceleration of a bobsled turn on an ideally banked curve

Problem:

(a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?

(b) Calculate the centripetal acceleration.

(c) Does this acceleration seem large to you?

Solution:

Part A

For ideally banked curved, the ideal banking angle is given by the formula $\displaystyle \tan \theta = \frac{v^2}{rg}$ . We can solve for $r$ in terms of all the other variables, and we should come up with

r = \frac{v^2}{g \tan \theta}

We are given the following values:

ideal banking angle, $\displaystyle \theta = 75.0\ ^\circ$

linear speed, $\displaystyle v=30.0\ \text{m/s}$

acceleration due to gravity, $\displaystyle g=9.81\ \text{m/s}^2$

If we substitute all the given values into our formula for $r$ , we have

\begin{align*} r & = \frac{v^2}{g \tan \theta} \\ \\ r & = \frac{\left( 30.0\ \text{m/s} \right)^2}{\left( 9.81\ \text{m/s}^2 \right)\left( \tan 75^\circ \right)} \\ \\ r & = 24.5825\ \text{m} \\ \\ r & = 24.6\ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The radius of the ideally banked curve is approximately $24.6\ \text{m}$ .

Part B

The centripetal acceleration $a_c$ can be solved using the formula

a_c = \frac{v^2}{r}

Substituting the given values, we have

\begin{align*} a_c & = \frac{v^2}{r} \\ \\ a_c & = \frac{\left( 30.0\ \text{m/s} \right)^2}{24.5825\ \text{m}} \\ \\ a_c & = 36.6114\ \text{m/s}^2 \\ \\ a_c & = 36.6 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The centripetal acceleration is about $36.6\ \text{m/s}^2$ .

Part C

To know how large is the computed centripetal acceleration, we can compare it with that of acceleration due to gravity.

\frac{a_c}{g} = \frac{36.6114\ \text{m/s}^2}{9.81\ \text{m/s}^2} = 3.73

The computed centripetal acceleration is 3.73 times the acceleration due to gravity. That is $a_c = 3.73g$ .

This does not seem too large, but it is clear that bobsledders feel a lot of force on
them going through sharply banked turns!

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.

(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?

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.)

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.

The Speed of the Roller Coaster at the Top of the Loop

Problem:

Solution:

The Ideal Speed and the Minimum Coefficient of Friction in Icy Mountain Roads

Problem:

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).

(a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º.

(b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?

The centripetal acceleration of a large centrifuge as experienced in rocket launches and atmospheric reentries of astronauts

Problem:

A large centrifuge, like the one shown in Figure 6.34(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.

(a) At what angular velocity is the centripetal acceleration 10g if the rider is 15.0 m from the center of rotation?

Solution:

Part A

Part B

The radius and centripetal acceleration of a bobsled turn on an ideally banked curve

Problem:

(a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?

(b) Calculate the centripetal acceleration.

(c) Does this acceleration seem large to you?

Solution:

Part A

Part B

Part C

(a) At what angular velocity is the centripetal acceleration $10g$ if the rider is 15.0 m from the center of rotation?