Question

20 kg block is initially at rest on a rough horizontal

surface. A horizontal force of 75 N is required to set

the block in motion. After it is in motion, a horizontal

force of 60 N is required to keep the block moving with

constant speed. The coefficient of static friction is ?​

Answer 1

Coefficient of static friction μ = 0.3.

The weight of the block (W) = 20 Kg.

Horizontal force applied on the block to set it in to motion (F) = 75 N.

Force required to keep the block moving at constant speed (F₂) = 60 N.

Coefficient of static friction (μ) = $\frac{(f_s)_(max)}{R}$

Normal reaction (R) = mass × gravitational constant = mg.

g = 10 m/s²

(μ) = $\frac{(f_s)_(max)}{mg}$

applied force is equal to static frictional force (f_s) in case the applied force leave body at rest or in constant speed, so we consider the force required for constant speed (F₂).

F₂ = (f_s) = 60 N

μ = $\frac{(60)}{(20)(10)}$

μ = $\frac{(60)}{200}$

μ = $\frac{(6)}{20}$

μ = [tex]\frac{(3)}{10}[/tex

μ = 0.3

Answer 2

Answer:

this is the answer for the above question