A 200-N sled of slides down a frictionless hillside that rises at 37° above the horizontal.What is the magnitude of the force that the surface of the hill exerts on the sled?
A)200 N
B)120 N
C)160 N
D)150 N
E)0 N

Question

Quizplus · Accepted Answer

The force of gravity acting on the sled is parallel to the surface of the hill and can be resolved into two components - one perpendicular to the surface (the normal force) and one parallel to the surface (the force causing the sled to slide down the hill). Since there is no friction, the parallel component is the only force causing the sled to move, and its magnitude is given by F = mg sin(37°) = (200 N)(sin 37°) ≈ 120 N. The normal force, which is equal in magnitude and opposite in direction to the perpendicular component of the weight, is given by N = mg cos(37°) = (200 N)(cos 37°) ≈ 160 N. Therefore, the magnitude of the force that the surface of the hill exerts on the sled is equal to the normal force, which is approximately 160 N.

A 200-N Sled of Slides Down a Frictionless Hillside That