Question

A 29.0-kg block is initially at rest on a horizontal surface. A horizontal force of 71.0 N is required to set the block in motion, after which a horizontal force of 56.0 N is required to keep the block moving with constant speed.

Answer 1

Answer:

(a) $\mu_s=0.25$

(b) $\mu_k=0.20$

Explanation:

According to Newton's second law:

$\sum F_y:N=mg\\\sum F_x:F_a=F_f$

Recall that the frictional force is related jointly with the coefficient of friction and normal force $F_f=\mu N$. Replacing in the above equation, we get the coefficient of friction:

$F_a=\mu N=\mu mg\\\mu=\frac{F_a}{mg}$

(a) The  coefficient of static friction is related with the force required to set the block in motion:

$\mu_s=\frac{71N}{29kg*9.8\frac{m}{s^2}}\\\mu_s=0.25$

(b) The  coefficient of kinetic friction is related with the force required to keep the block moving with constant speed:

$\mu_k=\frac{56N}{29kg*9.8\frac{m}{s^2}}\\\mu_k=0.20$