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 7.37. 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 7.37 A rescue sled and victim are lowered down a steep slope.

Free body diagram of a rescue sled and victim are lowered down a steep slope.

Work: Work is the product of the component of force in the displacement direction and the magnitude of the displacement.

(a)

The friction force is given as,

\(\begin{aligned}f &= \mu N\\ &= \mu W\cos \left( {60^\circ } \right)\\ &= \mu mg\cos \left( {60^\circ } \right)\end{aligned}\)

Here, \(\mu \) is the coefficient of friction \(\left( {\mu = 0.1} \right)\), \(N\) is the normal reaction force between sled and slope, \(W\) is the weight of the rescue sled and victim, \(m\) is the mass of the rescue sled and victim \(\left( {m = 90{\rm{ kg}}} \right)\), and \(g\) is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\).

Putting all known values,

The work done by the friction force is,

\({W_f} = fd\)

Here,\(d\)is the displacement of the sled\(\left( {d = 30.0{\rm{ m}}} \right)\).

Putting all known values,

\(\begin{aligned}{W_f} &= \left( {44.1{\rm{ N}}} \right) \times \left( {30.0{\rm{ m}}} \right)\\ &= 1323{\rm{ J}}\end{aligned}\)

Therefore, the required work is done by friction is \(1323{\rm{ J}}\).

(b)

The equation for the tension in the rope is,

\(\begin{aligned}T + f = mg\sin \left( {60^\circ } \right)\\T = mg\sin \left( {60^\circ } \right) - f\end{aligned}\)

Puttingall known values,

The work done by the rope is,

\(\begin{aligned}{W_r} &= Td\\ &= \left( {{\rm{719}}{\rm{.73 N}}} \right) \times \left( {30.0{\rm{ m}}} \right)\\ \approx 21592{\rm{ J}}\end{aligned}\)

Therefore, the required work is done by the rope is \(21591.9{\rm{ J}}\).

(c)

The height attained by the sled is,

\(h = d\sin \left( {60^\circ } \right)\)

Puttingall known values,

\(\begin{aligned}h &= \left( {30.0{\rm{ m}}} \right) \times \sin \left( {60^\circ } \right)\\ &= 25.98{\rm{ m}}\end{aligned}\)

The work done by the gravitational force is,

\({W_g} = - mgh\)

Puttingall known values,

\(\begin{aligned}{W_g} &= - \left( {90.0{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {25.98{\rm{ m}}} \right)\\ \approx - 22915{\rm{ J}}\end{aligned}\)

Therefore, the required work is done by the gravitational force is \( - 22915{\rm{ J}}\).

(d)

The total work is,

\(W = {W_f} + {W_r} + {W_g}\)

Puttingall known values,

\(\begin{aligned}W = \left( {1323{\rm{ J}}} \right) + \left( {21592{\rm{ J}}} \right) + \left( { - 22915{\rm{ J}}} \right)\\ = 0{\rm{ J}}\end{aligned}\)

Therefore, the required total work done is \(0{\rm{ J}}\).