Tag Archives: work done friction

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:

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 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*}