Question

A block of mass of 10 kg is kept on rough inclined plane is kept inside a lift. At t = 0, the system is set in motion, the lift accelerates upwards with 10 $m/s^2$. Find the work done by normal reaction on the block in one second with respect to lift. (m = 10 kg; $\theta ={37}^0$; $\mu =0.8$; $a_0=10m/s^2$)

Answer 1

Normal force $N=(ma+ma)\cos \theta$

Frictional force $f_s$

$f_s=\mu N$

$=\mu (ma+mg)\cos \theta$

Acceleration of block is $a_1$ then we have,

$(mg+ma)\sin \theta -f_s=m{a}_1$

$(mg+ma)\sin \theta$$-\mu (ma+mg)\cos \theta$ $=m{a}_1$

$a_1=\frac{(10\times 10+10\times 10)\frac{3}{5}-0.8(10\times 10+10\times 10)\frac{4}{5}}{10}$

$=12-0.8\times 16=-0.8$

Here, $a_1$ is positive means block do not move with respect to lift.

Hence work done by Normal reaction=0 w.r.t lift.