Question

A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s². What is the coefficient of kinetic friction between the block and the plane?

Answer 1

Let m be the mass of the body.



From the free body diagram,
R − mg = 0
(where R is the normal reaction force and g is the acceleration due to gravity)
⇒ R = mg (1)
Again ma − μ_kR = 0
(where μ_k is the coefficient of kinetic friction and a is deceleration)
or ma = μ_kR
From Equation (1),
ma = μ_kmg
⇒ a = μ_kg
⇒ 4 = μ_kg
$\Rightarrow {\mu }_{\mathrm{k}}=\frac{4}{g}=\frac{4}{10}=0.4$

Hence, the coefficient of the kinetic friction between the block and the plane is 0.4.