Question

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

Answer 1

Answer:

0.3826

Explanation:

The coefficient of static friction is given as :

μ = F/mg

= 75/ (20 × 9.8)

= 75/196

= 0.3826

Hence,

the coefficient of static friction is 0.3826.

Answer 2

Answer:-

• The coefficient of static friction is 0.375

Solution:-

Given:-

1. Mass of block (m)= 20 kg
2. Force required to set the block in motion (F) = 75N
3. g = 10m/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.