Question

1. Suppose a 10 N force is applied to the side of a 4.0 kg block that is sitting on a table what would be the minimum value of the coefficient of static friction in order for the block to remain motionless?

Answer 1

Answer:

$\mu= 0.25$

Explanation:

Given data

Force= 10N

mass= 4kg

r= 4*9.81

r= 39.24N

The expression for the force acting is expressed as

$F=\mu*r$

substitute

$\mu=F/r$

$\mu= 10/39.24$

$\mu= 0.25$