Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 8

Question

Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 85.0 kg, down a = 64.0° slope at constant speed, as shown in Figure 6.22. The coefficient of friction between the sled and the snow is 0.100. Figure 6.22 (a) How much work is done by friction as the sled moves 30.0 m along the hill? Incorrect: Your answer is incorrect. J (b) How much work is done by the rope on the sled in this distance? J (c) What is the work done by gravity on the sled? J (d) What is the total work done? J

Explanation / Answer

The force on the sled are
Friction force = uMgCos64 where u is coefficient of friction , M is mass of sled
Gravity force = Mg Sin64   
Tension of rope = T
Since the sled is moving with the constant speed therefore the equation according to second law.
MgSin64 - uMgCos64 = T
T = 749.179 N
Now the work done
Friction = uMgCos64*30 = 1096.607 J
Rope = T*30 = 749.179*30 = 22479.3 J
Gravity = Mgsin64*30 = 22483.78 J
Now total = Gravity + Rope - frcition = 43866.475 J

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