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Homework Physics? ?
Suppose the ski patrol lowers a rescue sled carrying an injured skiier, with a combined mass of 82.5 kg, down a 60.0° slope at constant speed, as shown in the Figure. The coefficient of kinetic friction between the sled and the snow is 0.100.
A.) How much work, in joules, is done by friction as the sled moves 29.5 m along the hill?
B.) How much work, in joules, is done by the rope on the sled over this distance?
C.) What is the work, in joules, done by the gravitational force on the sled?
D.) What is the net work done on the sled, in joules?
1 Answer
- ?Lv 73 days agoFavorite Answer
A) The friction force is against the direction of motion, so
Wfriction = -µ*m*g*d*cosΘ
Wfriction = -0.100*82.5kg*9.81m/s²*29.5m*cos60.0º
Wfriction = -1190 J ..... to 3 significant digits
B) The rope tension is against the direction of motion, so
Wtension = -m*g*d*sinΘ - Wfriction
(or longhand = -m*g*d*(sinΘ - µ*cosΘ)
Wtension = -82.5kg * 9.81m/s² * 29.5m * sin60.0º - -1190J
Wtension = -19500 J ..... to 3 significant digits
C) Wgravity = mgh = mgd*sinΘ
Wgravity = 82.5kg * 9.81m/s² * 29.5m * sin60.0º
Wgravity = 20700 J ... to 3 significant digits
D) net work = 0 -- "constant velocity" means no change in KE
