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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:

We are still uploading the solution to this problem. Please bear with us. If you need it, you can purchase the complete solution manual here.


College Physics by Openstax Chapter 6 Problem 28

Riding a Bicycle in an Ideally Banked Curve


Problem:

Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 6.33. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).

(a) Show that \theta (as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway—that is, \theta = \tan ^{-1} \left( v^2/rg \right)

(b) Calculate \theta for a 12.0 m/s turn of radius 30.0 m (as in a race).

Figure 6.33 A bicyclist negotiating a turn on level ground must lean at the correct angle—the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system, while its horizontal component must supply the centripetal force. This process produces a relationship among the angle θ, the speed v, and the radius of curvature r of the turn similar to that for the ideal banking of roadways.

Solution:

Part A

Let us redraw the given forces in a free-body diagram with their corresponding components.

The force N and F_c are the vertical and horizontal components of the force F.

If we take the equilibrium of forces in the vertical direction (since there is no motion in the vertical direction) and solve for F, we have

\begin{align*}
\sum F_y & = 0 \\ \\
F \cos \theta - mg & = 0 \\ \\
F \cos \theta & = mg \\ \\
F & = \frac{mg}{\cos \theta}  \quad \quad  & \color{Blue}  \small \text{Equation 1}
\end{align*}

If we take the sum of forces in the horizontal direction and equate it to mass times the centripetal acceleration (since the centripetal acceleration is directed in this direction), we have

\begin{align*}
\sum F_x & = ma_c \\ \\
F \sin \theta  & = m a_c \\ \\
F \sin \theta  & = m \frac{v^2}{r}   \quad \quad  & \color{Blue}  \small \text{Equation 2}
\end{align*}

We substitute Equation 1 to Equation 2.

\begin{align*}
F \sin \theta  & = m \frac{v^2}{r} \\ \\
\frac{mg}{\cos \theta} \sin \theta & = m \frac{v^2}{r} \\ \\
mg \frac{\sin \theta}{\cos \theta} & =m \frac{v^2}{r} \\ \\
\end{align*}

We can cancel m from both sides, and we can apply the trigonometric identity \displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}. We should come up with

\begin{align*}
g \tan \theta & = \frac{v^2}{r} \\ \\
\tan \theta & = \frac{v^2}{rg} \\ \\
\theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are given the following values:

  • linear velocity, v = 12.0\ \text{m/s}
  • radius of curvature, r=30.0\ \text{m}
  • acceleration due to gravity, g = 9.81\ \text{m/s}^2

We substitute the given values to the formula of \theta we solve in Part A.

\begin{align*}
\theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \\ \\
\theta & = \tan ^ {-1} \left[ \frac{\left( 12.0\ \text{m/s} \right)^2}{\left( 30.0\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\
\theta & = 26.0723 ^\circ \\ \\
\theta & = 26.1 ^\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}

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} \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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College Physics by Openstax Chapter 6 Problem 21

Centripetal acceleration of an amusement park ride shaped like a Viking ship


Problem:

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity. The speed at the bottom of the arc is 23.4 m/s.

(a) What is the centripetal acceleration at the bottom of the arc?

(b) Draw a free-body diagram of the forces acting on a rider at the bottom of the arc.

(c) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.

(d) Discuss whether the answer seems reasonable.


Solution: