91Ó°ÊÓ

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 \(0{\rm{ J}}\).