Question

A 25kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in motion. Once the chair is in motion, a 127 N horizontal force keeps it moving at a constant velocity. Find the coefficient of static friction between the chair and the floor.

Answer 1

Answer:

$\mu_k=0.51$  

Explanation:

Given that

Mass , m = 25 kg

We know that when body is in rest condition then static friction force act on the body and when body is in motion the kinetic friction force act on the body .That is why these two forces are given as follows

Static friction force ,fs= 165 N

Kinetic friction force ,fk = 127 N

If the body is moving with constant velocity ,it means that acceleration of that body is zero and all the forces are balanced.

Lets take coefficient of kinetic friction  = μk

The kinetic friction is given as follows

fk = μk  m g

Now by putting the values

127 = μk x 25 x 9.81

$\mu_k=\dfrac{127}{25\times 9.81}$

$\mu_k=0.51$

Therefore the value of coefficient of kinetic friction will be 0.51