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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*}

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:

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 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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Tag Archives: College Physics by Openstax

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.

College Physics by Openstax Chapter 7 Problem 5

College Physics by Openstax Chapter 7 Problem 4

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?

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.)

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.

College Physics by Openstax Chapter 6 Problem 31

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

Problem:

Solution:

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?