Question

A block slides from rest from the top of an inclined plane 8 m long which is inclined 35° with  the horizontal. If the coefficient of kinetic friction is 0.20, determine how long it will take the block to reach the bottom of the plane.

Answer 1

Let mass of block = m kg

Taking acceleration due to gravity = 9.8 m/s²

Coefficient of friction, $\mu=0.20$

Length of inclined plane = 8 m

From the above figure, balancing the forces perpendicular to the inclined surface,

$N=mgcos\theta\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(1)$

Balancing the forces along the inclined plane

$ma=mgsin\theta -\mu N\\\Rightarrow ma=mgsin\theta-\mu mgcos\theta\space\space\space\space\space\space\space\space\space\space\space\space\space from(1)\\\Rightarrow a=g(sin\theta-\mu cos\theta)$

From Newton's laws of motion,

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow t=\sqrt{\dfrac{2s}{a}}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space Since\space u=0\space m/s^2$

$\Rightarrow t=\sqrt{\dfrac{2\times8}{9.8(sin35\degree-0.20cos35\degree)}}\\\Rightarrow t=1.996\space s\approx2\space s$