Question

A 108 kg clock initially at rest on a horizontal floor requires a 639 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 521 N keeps it moving with a constant velocity.

Answer 1

Answer:

$\mu_s=0.60$

Explanation:

It is given that,

Mass of the clock, m = 108 kg

Force acting on it when it is in motion, $F=639\ N$

After the clock is in motion, a horizontal force of 521 N keeps it moving with a constant velocity, F' = 521 N

It is assumed to find the coefficient of between the clock and the floor. The force of friction is given by :

$F=\mu_smg$

$\mu_s=\dfrac{F}{mg}$

$\mu_s=\dfrac{639\ N}{108\ kg\times 9.8\ m/s^2}$

$\mu_s=0.60$

So, the coefficient of static friction between the clock and the floor is 0.6. Hence, this is the required solution.