Question

A block of mass 10kg is released on rough inclined plane. Block start descending with acceleration 2m\s². Kinetic friction force acting on block is

Answer 1

30N

Explanation:

Given values

Mass, m = 10kg

acceleration, a =  2m\s²

gravity, g = 10m\s²

$\theta = 30^o$

According to opposing friction,

$mg\sin\theta - f = m a$

$f = mg\sin\theta - ma$

$= 10 \times 10 \times \frac{1}{2} - 10 \times2$

$= 50 -20 = 30$

Thus, Kinetic friction force is 30N