A polar explorer pulls a sled on dry snow (the coefficient of kinetic friction i

Question

A polar explorer pulls a sled on dry snow (the coefficient of kinetic friction is 0.12 for sled on snow). The rope attached to the sled pulls at an angle of 9° above the horizontal and the explorer pulls with with 600 N of force. The sled has a total mass 71 kg. While the explorer pulls the sled for a horizontal distance of 41 m (before he needs to take a break),
a) What is the mechanical work done by the explorer (i.e. the rope) on the sled?
.

b) What is the magnitude of the normal force on the sled?


c) What is the mechanical work done by the normal force on the sled?


d) What is the magnitude of the force of friction on the sled?


e) What is the mechanical work done by the force of friction on the sled?

Explanation / Answer

The rope makes an angle of 9 degree above horizontal. The explorer pulls with 600 N force. Hence the component of force that is normal to the sled is = 600.sin(9) = 93.86 N. The work done by the normal force on the sled is = 93.86*41 = 3848.26 J. The weight of the sled is 71 Kg i.e. it is pulled by gravity by a weight of 710 N. The coefficient of kinetic friction determines the force of friction. The frictional force is the coefficient of friction times weight of sled = 0.12 * 710 = 85.2 N. This force opposes horizontal motion. The sled moves for 41 m. Hence the mechanical work done by friction on the sled is 85.2*41 = 3493.2 J.

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