Tag Archives: Solution Manual for College Physics by Openstax

College Physics by Openstax Chapter 8 Problem 8

A car moving at 10.0 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70.0 kg.

Solution:

Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes. In symbols, Newton’s second law of motion is defined to be

\text{F}_\text{net} = \frac{\Delta \textbf{p}}{\Delta t}

Moreover, Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity.

\textbf{p} = m \textbf{v}

So, the Newton’s second law becomes

\text{F}_\text{net} = \frac{m\Delta\textbf{v}}{\Delta t}

Substituting the given values, we have

\begin{align*} \text{F}_\text{net} = & \frac{m \Delta\textbf{v}}{\Delta t} \\ \text{F}_\text{net} = & \frac{\left( 70\ \text{kg} \right)\left( 10\ \text{m/s} \right)}{0.26\ \text{s}} \\ \text{F}_\text{net} = & 2692.3077 \ \text{N} \\ \text{F}_\text{net} = & 2.69 \times 10^{3}\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

College Physics Chapter 8 Problem 7

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600.0 m/s in a time of 2.00 ms (milliseconds)?

Solution:

Impulse, or change in momentum, equals the average net external force multiplied by the time this force acts:

\Delta \textbf{p}=F_{\text{net}} \Delta t

The unknown in this problem is the average force. We can solve the formula for force in terms of the other quantities.

F_{\text{net}} = \frac{\Delta \textbf{p}}{\Delta t}

We are given the following values:

\begin{align*} \Delta \textbf{p} & = m \Delta v = \left( 0.0300\ \text{kg} \right)\left( 600.0\ \text{m/s} \right) = 18.00\ \text{kg}\cdot \text{m/s} \\ \Delta t& = 2.00 \times 10^{-3}\ \text{s} \end{align*}

Substituting these given values, we can solve for the unknown.

\begin{align*} F_{\text{net}} & = \frac{\Delta \textbf{p}}{\Delta t} \\ F_{\text{net}} & = \frac{18.00\ \text{kg}\cdot \text{m/s}}{2.00 \times 10^{-3}\ \text{s}}\\ F_{\text{net}} & = 9000\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

College Physics by Openstax Chapter 8 Problem 3

(a) At what speed would a 2.00×10⁴-kg airplane have to fly to have a momentum of 1.60×10⁹ kg⋅m/s (the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of 60.0 m/s? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship.

Solution:

Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum $\textbf{p}$ is defined as

\textbf{p}=m \textbf{v},

where $m$ is the mass of the system and $\textbf{v}$ is its velocity.

Part A. The Speed of an Airplane Given its Momentum

The airplane has the following quantities given:

\begin{align*} m_{\text{airplane}} & = 2.00 \times 10^{4}\ \text{kg} \\ \textbf{p}_{\text{airplane}} & = 1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s} \end{align*}

Using the formula of momentum, we can solve for the velocity in terms of the other variables. We can then substitute the given quantities to solve for the velocity of the airplane.

\begin{align*} \textbf{p}_{\text{airplane}} & = m_{\text{airplane}} \textbf{v}_{\text{airplane}} \\ \textbf{v}_{\text{airplane}} & = \frac{\textbf{p}_{\text{airplane}}}{m_{\text{airplane}}} \\ \textbf{v}_{\text{airplane}} & = \frac{1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s}}{2.00 \times 10^{4}\ \text{kg}} \\ \textbf{v}_{\text{airplane}} & = 80000\ \text{m}/\text{s} \\ \textbf{v}_{\text{airplane}} & = 8.00 \times 10^{4}\ \text{m}/\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. The Airplane’s Momentum When it is Taking Off

In this case, we are given the following properties of an airplane taking off:

\begin{align*} m_{\text{airplane}} & = 2.00 \times 10^{4}\ \text{kg} \\ \textbf{v}_{\text{airplane}} & = 60.0\ \text{m}/\text{s} \\ \end{align*}

The momentum of the airplane at this instance is calculated as

\begin{align*} \textbf{p}_{\text{airplane}} & = m_{\text{airplane}} \textbf{v}_{\text{airplane}} \\ \textbf{p}_{\text{airplane}} & = \left( 2.00 \times 10^{4}\ \text{kg} \right)\left( 60.0\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{airplane}} & = 1200000\ \text{kg} \cdot \text{m}/\text{s} \\ \textbf{p}_{\text{airplane}} & = 1.20 \times 10^{6}\ \text{kg} \cdot \text{m}/\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

Since the momentum of the airplane is 3 orders of magnitude smaller than the ship, the ship will not recoil very much. The recoil would be -0.01 m/s, which is probably not noticeable.

College Physics by Openstax Chapter 8 Problem 2

(a) What is the mass of a large ship that has a momentum of 1.60×10⁹ kg⋅m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.

Solution:

Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum $\textbf{p}$ is defined as

\textbf{p}=m \textbf{v},

where $m$ is the mass of the system and $\textbf{v}$ is its velocity.

Part A. The Mass of a Large Ship Given its Momentum

For this part, we are given the following quantities of the ship:

\begin{align*} \textbf{p}_{\text{ship}} & = 1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s} \\ \textbf{v}_{\text{ship}} & = 48.0\ \text{km}/\text{h} = 13.3333\ \text{m}/\text{s} \end{align*}

From the formula of momentum, we can solve for $m$ in terms of $\textbf{p}$ and $\textbf{v}$ . Then we can substitute the given values to solve for the mass of the ship.

\begin{align*} \textbf{p}_{\text{ship}} & = m_{\text{ship}} \textbf{v}_{\text{ship}} \\ m_{\text{ship}} & = \frac{\textbf{p}_{\text{ship}}}{\textbf{v}_{\text{ship}}} \\ m_{\text{ship}} & = \frac{1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s}}{13.3333\ \text{m}/\text{s}} \\ m_{\text{ship}} & = 120000000\ \text{kg} \\ m_{\text{ship}} & = 1.20 \times 10^{8} \ \text{kg}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. Comparing the Momentum of a Large Ship with an Artillery Shell

The artillery shell has a mass of 1100 kilograms and a speed of 1200 m/s. Therefore, its momentum is

\begin{align*} \textbf{p}_{\text{shell}} & = m_{\text{shell}} \textbf{v}_{\text{shell}} \\ \textbf{p}_{\text{shell}} & = \left( 1100\ \text{kg} \right)\left( 1200\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{shell}} & = 1320000\ \text{kg} \cdot \text{m}/\text{s} \\ \textbf{p}_{\text{shell}} & = 1.32 \times 10^{6}\ \text{kg} \cdot \text{m}/\text{s} \end{align*}

Comparing the momentum of the large ship and the shell, we have.

\begin{align*} \frac{\textbf{p}_{\text{ship}}}{\textbf{p}_{\text{shell}}} & = \frac{1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s}}{1.32 \times 10^{6}\ \text{kg} \cdot \text{m}/\text{s}} \\ \frac{\textbf{p}_{\text{ship}}}{\textbf{p}_{\text{shell}}} & = 1212.1212 \\ \textbf{p}_{\text{ship}} & = 1212.1212\ \textbf{p}_{\text{shell}} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The ship has a momentum that is about 1200 times the momentum of the artillery shell.

College Physics by Openstax Chapter 8 Problem 1

(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of 7.50 m/s. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of 600 m/s. (c) What is the momentum of the 90.0-kg hunter running at 7.40 m/s after missing the elephant?

Solution:

Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum $\textbf{p}$ is defined as

\textbf{p}=m \textbf{v},

where $m$ is the mass of the system and $\textbf{v}$ is its velocity.

Part A. The Momentum of the Elephant

We are given the mass and the velocity of the elephant, so we can just directly substitute these values in the formula for momentum.

\begin{align*} \textbf{p}_{\text{elephant}} & = m_{\text{elephant}} \textbf{v}_{\text{elephant}} \\ \textbf{p}_{\text{elephant}} & = \left( 2000\ \text{kg} \right)\left( 7.50\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{elephant}} & = 15000\ \text{kg} \cdot \text{m}/\text{s} \\ \textbf{p}_{\text{elephant}} & = 1.50 \times 10 ^{4} \ \text{kg} \cdot \text{m}/\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. Comparing the momentum of the elephant in part A with the momentum of a tranquilizer

First, we need to calculate the momentum of the tranquilizer.

\begin{align*} \textbf{p}_{\text{tranquilizer}} & = m_{\text{tranquilizer}} \textbf{v}_{\text{tranquilizer}} \\ \textbf{p}_{\text{tranquilizer}} & = \left( 0.0400\ \text{kg} \right)\left( 600\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{tranquilizer}} & = 24\ \text{kg} \cdot \text{m}/\text{s} \end{align*}

Now, we can compare their momentums.

\begin{align*} \frac{\textbf{p}_{\text{elephant}}}{\textbf{p}_{\text{tranquilizer}}} & = \frac{15000\ \text{kg} \cdot \text{m}/\text{s}}{24\ \text{kg} \cdot \text{m}/\text{s} } \\ \frac{\textbf{p}_{\text{elephant}}}{\textbf{p}_{\text{tranquilizer}}} & = 625 \\ \textbf{p}_{\text{elephant}} & = 625\ \textbf{p}_{\text{tranquilizer}} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The momentum of the elephant is 625 times larger than the momentum of the tranquilizer.

Part C. The Momentum of a Hunter Running after missing the elephant

\begin{align*} \textbf{p}_{\text{hunter}} & = m_{\text{hunter}} \textbf{v}_{\text{hunter}} \\ \textbf{p}_{\text{hunter}} & = \left( 90.0\ \text{kg} \right)\left( 7.40\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{hunter}} & = 666\ \text{kg} \cdot \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

College Physics by Openstax Chapter 7 Problem 10

(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates.

Solution:

The translational kinetic energy of an object of mass $m$ moving at speed $v$ is $KE=\frac{1}{2}mv^{2}$ .

Part A. The Velocity of the Elephant to have the same Kinetic Energy as the Sprinter

First, we need to solve for the kinetic energy of the sprinter.

\begin{align*} KE_{\text{sprinter}} & = \frac{1}{2} \left( 65.0\ \text{kg} \right)\left( 10.0\ \text{m}/\text{s} \right)^{2} \\ KE_{\text{sprinter}} & = 3250\ \text{J} \end{align*}

Then, we need to equate this to the kinetic energy of the elephant with the velocity as the unknown.

\begin{align*} KE_{\text{elephant}} & = KE_{\text{sprinter}} \\ \frac{1}{2}\left( 3000\ \text{kg} \right) v^{2} & = 3250\ \text{J} \\ 1500 v^{2} & = 3250 \\ v^{2} & = \frac{3250}{1500} \\ v & = \sqrt[]{\frac{3250}{1500}} \\ v & = 1.4720\ \text{m}/\text{s} \\ v & = 1.47\ \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. How Larger Energies Needed for the Movement of Larger Animals would Relate to Metabolic Rates

If the elephant and the sprinter accelerate to a final velocity of 10.0 m/s, then
the elephant would have a much larger kinetic energy than the sprinter.
Therefore, the elephant clearly has burned more energy and requires a faster
metabolic output to sustain that speed. $\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)$

College Physics by Openstax Chapter 7 Problem 9

Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h.

Solution:

The translational kinetic energy of an object of mass $m$ moving at speed $v$ is $KE=\frac{1}{2}mv^{2}$ .

The Kinetic Energy of the Truck

For the truck, we are given the following:

\begin{align*} m & = 20 000\ \text{kg} \\ v & = 110\ \frac{\text{km}}{\text{hr}} \times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{hr}}{3600\ \text{s}} = 30.5556\ \text{m}/\text{s} \end{align*}

Substitute these values to compute for the kinetic energy of the truck.

\begin{align*} KE_{t} & = \frac{1}{2} mv^{2} \\ KE_{t} & = \frac{1}{2} \left( 20 000\ \text{kg} \right) \left( 30.5556\ \text{m}/\text{s} \right)^{2} \\ KE_{t} & = 9 336 446.9136\ \text{J} \\ KE_{t} & = 9.34 \times 10^{6} \ \text{J} \end{align*}

The Kinetic Energy of the Astronaut

For the astronaut, we have the following given values

\begin{align*} m & = 80\ \text{kg} \\ v & = 27 500\ \frac{\text{km}}{\text{hr}} \times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{hr}}{3600\ \text{s}} = 7638.8889\ \text{m}/\text{s} \end{align*}

The kinetic energy of the astronaut is calculated as

\begin{align*} KE_{a} & = \frac{1}{2} mv^{2} \\ KE_{a} & = \frac{1}{2} \left( 80\ \text{kg} \right) \left( 7638.8889\ \text{m}/\text{s} \right)^{2} \\ KE_{a} & = 2 334 104 945 .0617\ \text{J} \\ KE_{a} & = 2.33 \times 10^{9} \ \text{J} \end{align*}

Comparing the Kinetic Energies of the truck and the astronaut

\begin{align*} \frac{KE_{a}}{KE_{t}} & = \frac{2 334 104 945 .0617\ \text{J}}{9 336 446.9136\ \text{J}} \\ \frac{KE_{a}}{KE_{t}} & = 250 \\ KE_{a} & = 250\ KE_{t} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The kinetic energy of the astronaut is 250 times larger than the kinetic energy of the truck.

College Physics by Openstax Chapter 7 Problem 8

Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60.0º slope at constant speed, as shown in Figure 7.34. The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done?

**Figure 7.34** A rescue sled and victim are lowered down a steep slope.

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 Work Done by the Friction on the Sled

First, let us calculate the magnitude of the friction force, $F_{f}$ . We can do this using the formula,

f= \mu _{s} N

where $f$ is the friction force, $\mu _{s}$ is the coefficient of static friction, and $N$ is the normal force directed perpendicular to the surface as shown in the free-body diagram below.

a victim resting on a rescue sled while being lowered at a constant speed by ski patrols. the total mass is 90 kilograms and the slope is 60 degrees, and the coefficient of friction is 0.100.

Let us solve for the magnitude of the normal force, $N$ , by summing up forces in the $y$ -direction and equating it to zero, since the body is in equilibrium (moving at constant speed).

\begin{align*} \sum F_{y} & = 0 \\ N - W \cos 60^{\circ} & = 0 \\ N - mg \cos 60^{\circ} & = 0 \\ N - \left( 90\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right) \cos 60^{\circ} & = 0 \\ N -441\ \text{N} & = 0 \\ N & = 441\ \text{N} \end{align*}

Now that we solved the normal force to be 441 newtons, we can now solve for the value of the frictional force, $f$ .

\begin{align*} f & = \mu _{s} N \\ f & = 0.100 \left( 441\ \text{N} \right) \\ f & = 44.1\ \text{N} \end{align*}

We can now substitute this value in the formula for work to solve for the work done by the friction force to the sled. We should also note that the friction force is against the direction of motion making the friction force and the displacement acting in opposite directions. This means that $\theta = 180^{\circ}$ .

\begin{align*} W_{f} & =fd \cos \theta \\ W_{f} & = \left( 44.1\ \text{N} \right)\left( 30.0\ \text{m} \right) \cos 180^{\circ }\\ W_{f} & = -1323\ \text{N} \cdot \text{m} \\ W_{f} & = -1323\ \text{J} \\ W_{f} & = -1.32 \times 10^{3} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. The Work Done by the Rope on the Sled

Using the same free-body diagram, we can solve for the magnitude of the force on the rope, $T$ . The symbol $T$ is used as this is a tension force from the rope.

A rescue sled and a victim being lowered down, having a total mass of 90 kilogram, down a 60 degree slope with a coefficient of friction of 0.001.

Taking the sum of forces in the $x$ -direction and equating it to zero.

\begin{align*} \sum F_{x} & = 0 \\ T + f - W \cos 30^{\circ } & = 0 \\ T + f - mg \cos 30^{\circ } & = 0 \\ T + 44.1\ \text{N} -\left( 90\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right) \cos 30^{\circ } & = 0 \\ T -719.7344\ \text{N} & = 0 \\ T & = 719.7344\ \text{N} \\ \end{align*}

Now, we can substitute this value to the formula of work. Note that the direction of motion is still opposite the direction of the force.

\begin{align*} W_r & =Td \cos \theta \\ W_r & = \left( 719.7344\ \text{N} \right)\left( 30.0\ \text{m} \right) \cos 180^{\circ }\\ W_r & = -21592.032\ \text{N} \cdot \text{m} \\ W_r & = -21592.032\ \text{J} \\ W_r & = -2.16 \times 10^{4} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C. The Work Done by the Gravitational Force on the Sled

The magnitude of the gravitational force can be easily calculated using the formula, $F_{g}=mg$ .

\begin{align*} F_{g} & = mg \\ F_{g} & = \left( 90\ \text{kg} \right)\left( 9.80\ \text{m}/\text{s}^2 \right) \\ F_{g} & = 882\ \text{kg} \cdot \text{m}/\text{s}^2 \\ F_{g} & = 882\ \text{N} \end{align*}

This is equivalent to the weight of the sled (and the victim). We can now substitute the weight of the sled and the displacement, knowing that the angle between these two quantities is $\theta = 30^{\circ}$ .

\begin{align*} W_g & = F_{g} d \cos \theta \\ W_g & = \left( 882\ \text{N} \right)\left( 30\ \text{m} \right) \cos 30^{\circ } \\ W_g & = 22915.0322\ \text{N} \cdot \text{m} \\ W_g & = 22915.0322\ \text{J} \\ W_g & = 2.29 \times 10^{4} \text{J}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D. The Total Work Done on the Sled

Since the sled moves at a constant speed, the net work done on the sled should be equal to zero. This is validated if we sum up all the works by each individual forces.

\begin{align*} W_{\text{net}} & = \sum W_{F} \\ W_{\text{net}} & = W_{f} + W_{r} +W_{g} \\ W_{\text{net}} & = -1323\ \text{J} + \left( -21592.032\ \text{J} \right)+22915.0322\ \text{J} \\ W_{\text{net}} & = 0\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

College Physics by Openstax Chapter 7 Problem 7

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.0º below the horizontal. (a) What is the work done on the cart by friction? (b) What is the work done on the cart by the gravitational force? (c) What is the work done on the cart by the shopper? (d) Find the force the shopper exerts, using energy considerations. (e) What is the total work done on the cart?

Solution:

W=Fd \cos \theta

Part A. The Work Done on the Cart by Friction

In this case, the friction opposes the motion. So, we have the following given values:

\begin{align*} F = & 35.0\ \text{N} \\ d = & 20.0\ \text{m} \\ \theta = & 180^{\circ } \\ \end{align*}

A shopper pusher a grocery cart showing that friction and displacement act in opposite directions.

The value of the angle $\theta$ indicates that $F$ and $d$ are directed in opposite directions. Substituting these values into the formula,

\begin{align*} W = & Fd \cos \theta \\ W = & \left( 35.0\ \text{N} \right)\left( 20.0\ \text{m} \right) \cos 180^{\circ } \\ W = & -700\ \text{N} \cdot \text{m} \\ W = & -700\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. Work Done on the Cart by the Gravitational Force

In this case, the gravitational force is directed downward while the displacement is horizontal as shown in the figure below.

A shopper pushes a grocery cart showing that displacement is horizontal while the gravitational force is downward.

We are given the following values:

\begin{align*} F = & mg\ \\ d = & 20.0\ \text{m} \\ \theta = & 90^{\circ } \\ \end{align*}

Substituting these values into the work formula, we have

\begin{align*} W = & Fd \cos \theta \\ W = & \left( \text{mg} \right)\left( 20.0\ \text{m} \right) \cos 90^{\circ } \\ W = & 0\ \text{N} \cdot \text{m} \\ W = & 0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

We can see that the gravitational force does not do any work on the cart because of the angle between the two quantities.

Part C. The Work on the Cart by the Shopper

Since we do not know the force exerted by the shopper, we are going to compute the work done by the shopper on the cart using the Work-Energy Theorem.

The work-energy theorem states that the net work $W_{\text{net}}$ on a system changes its kinetic energy. That is

W_{\text{net}} = \frac{1}{2}mv^{2}-\frac{1}{2}{mv_0} ^{2}

Now, we know that the shopper pushes the cart at a constant speed. This indicates that the initial and final velocities are equal to each other, making the net work $W_{\text{net}}$ is equal to zero.

W_{\text{net}} = 0

We also know that the total work done on the cart is the sum of the work done by the shopper and the friction force.

W_{\text{net}} = W_{\text{shopper}} +W_{\text{friction}}=0

This leaves us the final equation

\begin{align*} W_{\text{shopper}} + W_{\text{friction}} = & 0 \\ W_{\text{shopper}} + \left( -700\ \text{J} \right) = & 0 \\ W_{\text{shopper}} = & 700\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D. The force that the shopper exerts

In this case, the work of the shopper is directed 25 degrees below the horizontal while the displacement is still horizontal. This is depicted in the image below.

We are given the following values:

\begin{align*} W_{\text{shopper}} = & 700\ \text{J} \\ d = & 20.0\ \text{m} \\ \theta = & 25^{\circ } \\ \end{align*}

Substituting these values in the formula for work, we have

\begin{align*} W_{\text{shopper}} & = F_{\text{shopper}} d \cos \theta \\ F_{\text{shopper}} & = \frac{W_{\text{shopper}}}{d \cos \theta} \\ F_{\text{shopper}} & = \frac{700\ \text{J}}{\left( 20\ \text{m} \right)\cos 25^{\circ}} \\ F_{\text{shopper}} & = 38.6182\ \text{N} \\ F_{\text{shopper}} & = 38.6\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part E. The Net Work done on the cart

The net work done on the cart is the sum of work done by each of the forces, namely friction and shopper forces. That is,

\begin{align*} W_{\text{net}} & = W_{\text{shopper}} + W_{\text{friction}} \\ W_{\text{net}} & = 700\ \text{J} + \left( -700\ \text{J} \right) \\ W_{\text{net}} & = 0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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:

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*}

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College Physics by Openstax Chapter 8 Problem 8

A car moving at 10.0 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70.0 kg.

College Physics Chapter 8 Problem 7

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600.0 m/s in a time of 2.00 ms (milliseconds)?

College Physics by Openstax Chapter 8 Problem 3

College Physics by Openstax Chapter 8 Problem 2

(a) What is the mass of a large ship that has a momentum of 1.60×10⁹ kg⋅m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.

College Physics by Openstax Chapter 8 Problem 1

College Physics by Openstax Chapter 7 Problem 10

(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates.

College Physics by Openstax Chapter 7 Problem 9

Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h.

College Physics by Openstax Chapter 7 Problem 8

College Physics by Openstax Chapter 7 Problem 7

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.

A car moving at 10.0 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70.0 kg.

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600.0 m/s in a time of 2.00 ms (milliseconds)?

(a) What is the mass of a large ship that has a momentum of 1.60×109 kg⋅m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.

(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates.

Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h.

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) What is the mass of a large ship that has a momentum of 1.60×10⁹ kg⋅m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.