Question

A skier of mass 103 kg comes down a slopeof constant angle 32◦with the horizontal.What is the force on the skier parallel tothe slope? Neglect friction. The accelerationof gravity is 9.8 m/s2.Answer in units of N.

Answer 1

Answer:

534.9 N

Explanation:

The skier weight is his mass times gravitational acceleration g

W = mg = 103 * 9.8 = 1009.4 N

This weight can be divided into 2 components, one perpendicular and the other parallel to the 32-degree slope. The parallel component would equal to

$Wsin(32^0) = 1009.4sin(32^0) = 534.9 N$