Tag Archives: work done on a rescue sled

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 WW that a force FF does on an object is the product of the magnitude FF of the force, times the magnitude dd of the displacement, times the cosine of the angle θ\theta between them. In symbols,

W=Fdcos⁡θW=Fd \cos \theta

Part A. The Work Done by the Friction on the Sled

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

f=μsNf= \mu _{s} N

where ff is the friction force, μs\mu _{s} is the coefficient of static friction, and NN 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, NN, by summing up forces in the yy-direction and equating it to zero, since the body is in equilibrium (moving at constant speed).

∑Fy=0N−Wcos⁡60∘=0N−mgcos⁡60∘=0N−(90 kg)(9.80 m/s2)cos⁡60∘=0N−441 N=0N=441 N\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, ff.

f=μsNf=0.100(441 N)f=44.1 N\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 θ=180∘\theta = 180^{\circ}.

Wf=fdcos⁡θWf=(44.1 N)(30.0 m)cos⁡180∘Wf=−1323 N⋅mWf=−1323 JWf=−1.32×103 J  (Answer)\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, TT. The symbol TT 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 xx-direction and equating it to zero.

∑Fx=0T+f−Wcos⁡30∘=0T+f−mgcos⁡30∘=0T+44.1 N−(90 kg)(9.80 m/s2)cos⁡30∘=0T−719.7344 N=0T=719.7344 N\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.

Wr=Tdcos⁡θWr=(719.7344 N)(30.0 m)cos⁡180∘Wr=−21592.032 N⋅mWr=−21592.032 JWr=−2.16×104 J  (Answer)\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, Fg=mgF_{g}=mg.

Fg=mgFg=(90 kg)(9.80 m/s2)Fg=882 kg⋅m/s2Fg=882 N\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 θ=30∘\theta = 30^{\circ}.

Wg=Fgdcos⁡θWg=(882 N)(30 m)cos⁡30∘Wg=22915.0322 N⋅mWg=22915.0322 JWg=2.29×104J  (Answer)\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.

Wnet=∑WFWnet=Wf+Wr+WgWnet=−1323 J+(−21592.032 J)+22915.0322 JWnet=0 J  (Answer)\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*}