Question

A block of mass 1 kg is kept on incline plane inside lift as shown in figure. If lift starts accelerating up at 2 m/$s^2$ from rest then work done by frictional force in 1 second is (in J)
( g = 10 m/$s^2$)


Friction between block and plane is 0.8.

Answer 1

FBD of block w.r.t lift

$f_l=\mu N=\mu m(a+g)cos\theta =0.4\sqrt{3}(a+g)$ $\{\theta ={30}^{\circ }\}$
Since $f_l>m(a+g)sin\theta =0.5(a+g)$
$f=m(a+g)sin\theta =6N$
Work $=fsin\theta \times \frac{1}{2}\times a\times 1^2=3$ J