Tag Archives: College Physics Complete Solution Manual

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

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*}

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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*}

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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!

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College Physics by Openstax Chapter 6 Problem 26

The Ideal Speed on a Banked Curve

Problem:

What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?

Solution:

The formula for the ideal speed on a banked curve can be derived from the formula of the ideal angle. That is, starting from $\tan \theta = \frac{v^2}{rg}$ , we can solve for $v$ .

v = \sqrt{rg \tan \theta}

For this problem, we are given the following values:

radius of curvature, $r=100\ \text{m}$
acceleration due to gravity, $g=9.81\ \text{m/s}^2$
banking angle, $\theta = 20.0 ^\circ$

If we substitute the given values into our 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) \left( \tan 20.0 ^\circ \right) } \\ \\ v = & 18.8959\ \text{m/s} \\ \\ v = & 18.9\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The ideal speed for the given banked curve is about $18.9\ \text{m/s}$ .

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College Physics by Openstax Chapter 6 Problem 25

The ideal banking angle of a curve on a highway

Problem:

What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?

Solution:

The ideal banking angle (meaning there is no involved friction) of a car on a curve is given by the formula:

\theta = \tan^{-1} \left( \frac{v^2}{rg} \right)

We are given the following values:

radius of curvature, $\displaystyle r = 1.20\ \text{km} \times \frac{1000\ \text{m}}{1\ \text{km}} = 1200\ \text{m}$

linear velocity, $\displaystyle v=105\ \text{km/h}\times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3600\ \text{s}} = 29.1667\ \text{m/s}$

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

If we substitute these values into our formula, we come up with

\begin{align*} \theta & = \tan^{-1} \left( \frac{v^2}{rg} \right) \\ \\ \theta & = \tan^{-1} \left[ \frac{\left( 29.1667\ \text{m/s} \right)^2}{\left( 1200\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\ \theta & = 4.1333 ^\circ \\ \\ \theta & = 4.13 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The ideal banking angle for the given highway is about $4.13 ^\circ$ .

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College Physics by Openstax Chapter 6 Problem 24

Centripetal Force of a Rotating Wind Turbine Blade

Problem:

Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.

Solution:

We are given the following values:

radius, $r=100\ \text{m}$
angular velocity, $\omega = 0.5\ \text{rev/sec}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} = 3.1416\ \text{rad/sec}$
mass, $m=4\ \text{kg}$

F_c = m \frac{v^2}{r} \quad \text{or} \quad \ F_c = mr \omega^2

For this particular problem, we are going to use the formula $F_c = mr \omega^2$ . If we substitute the given values, we have

\begin{align*} F_c & =mr \omega^2 \\ \\ F_c & = \left( 4\ \text{kg} \right)\left( 100\ \text{m} \right)\left( 3.1416\ \text{rad/sec} \right)^2 \\ \\ F_c & = 3947.8602\ \text{N} \\ \\ F_c & = 4 \times 10^3\ \text{N}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The centripetal force on the end of the wind turbine blade is approximately $4 \times 10^3\ \text{N}$ .

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College Physics by Openstax Chapter 6 Problem 23

The centripetal force of a child riding a merry-go-round

Problem:

(a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center?

(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?

(c) Compare each force with her weight.

Solution:

Part A

We are given the following values: $m=22.0\ \text{kg}$ , $\omega = 40.0\ \text{rev/min}$ , and $r=1.25\ \text{m}$ . We are asked to solve for the centripetal force, $F_c$ .

Centripetal force $F_c$ is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity $v$ and has magnitude $F_c=m a_c$ , which can also be expressed as $F_c = m \frac{v^2}{r}$ or $F_c = m r \omega ^2$ . Basing from the given values, we are going to solve the problem using the formula

F_c = m r \omega ^2

First, we need to convert the angular velocity $\omega$ to $\text{rad/sec}$ for unit homogeneity.

40\ \text{rev/min} \times \frac{2\pi\ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} = 4.1888\ \text{rad/sec}

Now, we can substitute the given values into our formula.

\begin{align*} F_c & = m r \omega ^2 \\ \\ F_c & = \left( 22.0\ \text{kg}\right) \left( 1.25\ \text{m}\right) \left( 4.1888\ \text{rad/sec}\right)^2 \\ \\ F_c & = 482.5162\ \text{N} \\ \\ F_c & = 483\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Let us convert the angular velocity to radians per second.

3.00\ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}}=0.3142 \ \text{rad/sec}

Now, we can substitute the given values

\begin{align*} F_c & = m r \omega ^2 \\ \\ F_c & = \left( 22.0\ \text{kg}\right) \left( 8.00\ \text{m}\right) \left( 0.3142\ \text{rad/sec}\right)^2 \\ \\ F_c & = 17.3750\ \text{N} \\ \\ F_c & = 17.4\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

For the first centripetal force we solved in Part A,

\frac{F_c}{w} = \frac{483\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 2.24

The centripetal force is 2.24 times the weight of the child.

For the centripetal force we solved in Part B, we have

\frac{F_c}{w} = \frac{17.4\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 0.0806

The centripetal force is only about 8% of the child’s weight.

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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

The Ideal Speed on a Banked Curve

Problem:

What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?

Solution:

The ideal banking angle of a curve on a highway

Problem:

What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?

Solution:

Centripetal Force of a Rotating Wind Turbine Blade

Problem:

Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.

Solution:

The centripetal force of a child riding a merry-go-round

Problem:

(a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center?

(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?

(c) Compare each force with her weight.

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?