Question

In the figure, the wedge is pushed with an acceleration of $10\sqrt{3}m/s^2$. It is seen that the block starts climbing
upon the smooth inclined face of wedge. What will be the time taken by the block to reach the top?

• A. $\frac{2}{\sqrt{5}}s$
  
• B. $\frac{1}{\sqrt{5}}s$
  
• C. $\sqrt{5}s$
  
• D. $\frac{\sqrt{5}}{2}s$

Answer 1

The correct option is B $\frac{1}{\sqrt{5}}s$


$10\sqrt{3}cos{30}^{\circ }=10\sqrt{3}\frac{\sqrt{3}}{2}=15m/s^2$

$gsin{30}^{\circ }=5m/s^2$
$\therefore a=15-5=10m/s^2\therefore s=\frac{1}{2}a{t}^{2}1=\frac{1}{2}\left(10\right)t^{2}ort=\frac{1}{\sqrt{5}}s$