## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 2 - Problems - Page 68: 74

#### Answer

The normal force is $152.8~N$ The magnitude of the force of kinetic friction exerted on the crate is $61.1~N$ and this force is directed down the ramp at an angle of $30^{\circ}$ to the horizontal.

#### Work Step by Step

The normal force $F_N$ is the component of the crate's weight that is directed straight into the surface of the ramp. $F_N = mg~cos~\theta$ $F_N = (18.0~kg)(9.80~m/s^2)~cos~30^{\circ}$ $F_N = 152.8~N$ The normal force is $152.8~N$ Since the crate sliding up the ramp, the force of kinetic friction $F_f$ is exerted on the crate and it is directed down the ramp. We can find the force of kinetic friction: $F_f = F_N~\mu_k$ $F_f = mg~cos~\theta~\mu_k$ $F_f = (18.0~kg)(9.80~m/s^2)~cos~30^{\circ}~ (0.40)$ $F_f = 61.1~N$ The magnitude of the force of kinetic friction exerted on the crate is $61.1~N$ and this force is directed down the ramp at an angle of $30^{\circ}$ to the horizontal.

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