Question

A block of mass M = 4kg starts sliding downa rough inclined planc, the work done bygravitational force is twice the work done byfriction, then the kinetic energy of the block asit reaches the bottom of the inclined plane is​

Answer 1

Inclination be θ.

Height from starting of sliding be 'h' from ground.

Also, as is seen from diagram, s=h cosec θ  --------(1)

Then,

Work done by gravity= 2* work done by friction

Taking coefficient of kinetic friction to be μ and N as normal force of block on incline:

$mgh=muNs$

Using 1,

$mgh=\alpha *mg*cos(theta)*h*cosec(theta)\\cosec(theta)=\frac{sec(theta)}{2mu}$---------(2)

Using 2 in 1,

s=h*sec(θ)/2μ

Now, third eq of motion along incline:

v²=u²+2gs

v²=2g*hsec(θ)/2μ

v²=ghsec(θ)/μ             --------(3)

Using 3 we get:

Kinetic energy= $\frac{mv^2}2=\frac42 \frac{ghsec(theta)}{mu}$

Hence kinetic energy= 2gh*sec(θ)/μ