Question

A 20kg block is initially at rest .A 75N force is required to set the block in motion. After the motion starts, a force of 60N is applied to keep the block moving with constant speed.The coefficient of static friction is ??​

Answer 1

Answer:-

The coefficient of static friction is 0.375

Solution:-

Given:-

• Mass of block (m)= 20 kg

• Force required to set the block in motion (F) = 75N

• g = 9.8m/s²

To find:-

Coefficient of static friction (u ) = ??

So:-

We know that the force required to set the block in motion should be equal to limiting friction, thus we get

Let:-

Limiting static friction be f.

Then:-

=> f = uN

=> f = F

=> F = uN

=> 75N = u×mg

=> 75 /mg = u

=> 75/200 = u

=> u = 0.375

Therefore:-

The coefficient of static friction is 0.375.

Answer 2

Answer:

option 4

Explanation: