# Question #580ba

Jun 18, 2016

as below

#### Explanation:

Given

• $m \to \text{Mass of the object } = 100 k g$
• ${\mu}_{s} \to \text{Coefficient of static friction } = 0.60$
• ${\mu}_{s} \to \text{Coefficient of kinetic friction } = 0.645$

When placed on horizontal surface the normal reaction acting on the object is equal to its weight
so
$\text{Normal reaction } \left(N\right) = m g = 100 \times 9.8 N = 980 N$

$\text{Limiting value of static friction } {F}_{s} = {\mu}_{s} \times N = 0.6 \times 980 = 588 N$

So this amount force (588N) is needed to start moving

As the force of static friction is self adjusting below its limiting value,
application of 400N force on it being less than limiting value 588N the force of static friction will be equal to the force applied on it i.e. 400N.

When 650 N is applied on it ,it will move and the resisting force of
kinetic friction will be operative on it and this frictional force will be
$\text{Kinetic friction } \left({F}_{k}\right) = {\mu}_{k} \times N = 0.45 \times 980 N = 441 N$