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?

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.

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.

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*}