Question

A 20 kg sled is being pulled across a horizontal force at a constant velocity the pulling force has a magnitude of 80 N and it's directed at an angle of 30° above the horizontal determine the coefficient of kinetic friction

Answer 1

Answer:

0.58

Explanation:

Using the frictional force formula expressed as;

$F_f = \mu R$

$\mu$ is the coefficient of kinetic friction

R is the normal reaction

$\mu = \frac{F_f}{R}$

$F_f = Wsin \theta\\R = W cos\theta\\\mu = \frac{Wsin \theta}{W cos \theta}\\\mu = \frac{sin\theta}{cos \theta} \\\mu = tan \theta$

Given

$\theta = 30^0\\\mu = tan 30^0\\\mu = 0.58$

hence the coefficient of kinetic friction is 0.58