Complete Question
A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.
Answer:
The value for static friction is [tex]\mu_s = 0.60[/tex]
The value for static friction is [tex]\mu_k = 0.70[/tex]
Explanation:
From the question we are told that
The mass of the clock is [tex]m = 95 \ kg[/tex]
The first horizontal force is [tex]F _s = 560 \ N[/tex]
The second horizontal force is [tex]F _k = 650 \ N[/tex]
Generally the static frictional force is equal to the first horizontal force
So
[tex]F _s = m * g * \mu_s[/tex]
=> [tex]560 = 95 * 9.8 * \mu_s[/tex]
=> [tex]\mu_s = 0.60[/tex]
Generally the kinetic frictional force is equal to the second horizontal force
So
[tex]F _k = m * g * \mu_k[/tex]
[tex]650 = 95 * 9.8 * \mu_k[/tex]
[tex]\mu_k = 0.70[/tex]
