Question

A21.0-kg block is initially at rest on a horizontal surface. A horizontal force of73.0N is required to set the block in motion, after which a horizontal force of58.0N is required to keep the block moving with constant speed.(a) Find the coefficient of static friction between the block and the surface.(b) Find the coefficient of kinetic friction between the block and the surface.

Answer 1

Answer:

a) μs=0.34

b) μk=0.27

Explanation:

Given that

Mass , m = 21 kg

Force required to move the block ,F= 73 N

The force required to keep the block in motion ,F ' = 58 N

Coefficient of the static friction  :

The maximum value of the static friction force

fr= μs m g

At rest position

F=  fr

73 = μs x 21 x 10   (take g= 10 m/s²)

$\mu_s=\dfrac{73}{210}$

μs=0.34

Therefore the Coefficient of the static friction  will be 0.34.

Coefficient of the kinetic  friction  ;

Value of the kinetic friction force   ,fr= μk m g

F' = Fr

To move the block at constant speed

58 = μs x 21 x 10                 (take g= 10 m/s²)

$\mu_k=\dfrac{58}{210}$

μk=0.27

Therefore the Coefficient of the static friction  will be 0.27.