Question

A20 kg block is initially at a rough horizontal force of 75N is required to set the block in motion after it is in a motion, a horizontal force of 60N is required to keep a block moving with constant speed the coefficient of static friction

Answer 1

Force required to set the block in motion should be equal to limiting friction. Limiting friction is given by f = μN (where μ is coefficient of static friction and N is normal reaction. In this case N = mg)

F = f
F = μmg
75 = μ × 20 × 10
μ = 75/200
μ = 0.375

Coefficient of static friction is 0.375

Answer 2

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.