91Ó°ÊÓ

The rate at which velocity changes is referred to as acceleration.

Due to the component of gravitational force, the inclined object will move downward but this motion will be opposed by friction force which will act opposite to the inclined component of gravitation force.

The object will move downward due to the component of gravitational force so friction force will be up to the incline.

As block is at rest perpendicular to the inclined surface, therefore net force perpendicular to the inclined is zero so

$\mathrm{N}-\mathrm{mg}\sin \left(90-\mathrm{θ}\right)=0$

or $\mathrm{N}=\mathrm{mgcos}\left(\mathrm{θ}\right)$

So kinetic friction force,$\mathrm{f}={\mathrm{μ}}_{\mathrm{K}}\mathrm{N}={\mathrm{μ}}_{\mathrm{K}}\mathrm{mgcos}\left(\mathrm{θ}\right)$

Along the incline there is a net external force on the system as friction is of kinetic in nature so applying Newton’s second law along the inclined surface

$\mathrm{Net}\mathrm{force}=\mathrm{mass}×\mathrm{acceleration}$

$\mathrm{mgcos}\left(90-\mathrm{θ}\right)-\mathrm{f}=\mathrm{ma}$

or $\mathrm{mg}\sin \left(\mathrm{θ}\right)-{\mathrm{μ}}_{\mathrm{K}}\mathrm{mgcos}\left(\mathrm{θ}\right)=\mathrm{ma}$

Here mass is common on both the side so it will cancel out

Therefore, $\mathrm{a}=\mathrm{g}\sin \left(\mathrm{θ}\right)-{\mathrm{μ}}_{\mathrm{K}}\mathrm{g}\cos \left(\mathrm{θ}\right)$...............................(1)

$\mathrm{g}$is the same on both sides, and the object's acceleration down the incline is independent of mass.

$\mathrm{a}=\mathrm{g}\{\sin \left(\mathrm{θ}\right)-{\mathrm{μ}}_{\mathrm{K}}\cos \left(\mathrm{θ}\right)\}$

It is independent of mass.

when friction becomes negligibly small $({\mathrm{μ}}_{\mathrm{K}}=0)$then from equation (1)

$$\begin{gathered} \mathrm{a}=\mathrm{g}\sin \left(\mathrm{θ}\right)-0 \\ \mathrm{or}\mathrm{a}=\mathrm{g}\sin \left(\mathrm{θ}\right) \end{gathered}$$

It is the same acceleration like that of acceleration due to gravity on an inclined surface of angle$\mathrm{θ}$.