Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60.0â° slope at constant speed, as shown below. 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?

A. how much work is done by friction as the sled moves 30.0 m along the hill

We use the formula:

friction work = -µmgdcosΘ 

friction work = -0.100 * 90 kg * 9.8 m/s^2 * 30 m * cos 60

friction work = - 1,323 J

B. how much work is done by the rope on the sled in this distance?

We use the formula:

rope work = -mgd(sinΘ - µcosΘ) 
rope work = - 90 kg * 9.8 m/s^2 * 30 m (sin 60 – 0.100 * cos 60)

rope work = 21,592 J

C. what is the work done by the gravitational force on the sled?

We use the formula:

gravity work = mgdsinΘ 

gravity work = 90 kg * 9.8 m/s^2 * 30 m * sin 60

gravity work = 22,915 J

D. what is the total work done?

We add everything:

total work = - 1,323 J + 21,592 J + 22,915 J

total work = 43,184 J

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explanation: