Question

In the figure shown all the surfaces are frictionless, and mass of the block, $m=1\text{kg}$. The block and wedge are held initially at rest. Now wedge is given a horizontal acceleration of $10{\text{m/s}}^2$ by applying a force on the wedge, so that the block does not slip on the wedge. Then work done by the normal force in ground frame on the block in $\sqrt{3}$ seconds is (in $J$)

Answer 1

In ground frame horizontal acceleration of block is given as $10{\text{m/s}}^2$
So velocity of block is $v=at=10\sqrt{3}\text{m/s}$

Work energy theorem in ground frame gives
$W_{gravity}+W_{Normal}=\Delta K=\frac{m{v}^2}{2}$
$0+W_{Normal}=\frac{1}{2}\times 1\times (10\sqrt{3}{)}^2=150\text{J}$