A 20 kg block is initially at rest

Asked by admin @ 31/08/2021 in Science viewed by 286 People

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 ?

Answered by admin @ 31/08/2021

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

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