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 up on the smooth inclined face of wedge. What will be the time taken by the block to the 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

Answer: (B)

$\frac{1}{\sqrt{5}}s$

let the mass of the square block be m so the force acting on it is $F=mg$ in the vertical direction therefore the force has two components, one along the plane that is $mgcos{60}^o$ and one perpendicular to the plane $mgsin{60}^o$

the force along the plane is pushing the square block down and has deceleration of $\frac{F}{m}=\frac{mgcos{60}^o}{m}=gcos{60}^o=\frac{g}{2}=5m/s^2$

now for the block to climp up the acceleration of the triangular block should be greater than the deceleration of square block. the component of acceleration of rectangular block along the plane is $acos{30}^o=10\sqrt{3}\times \frac{\sqrt{3}}{2}=15m/s^2$ and perpendicular to the plane is $asin{30}^o$

therefore the resultant acceleration along the plane is $15-5=10m/s^2$

Time taken can be calculated by using newtons laws of motion.

$S=ut+\frac{1}{2}a{t}^2$

$1=0+\frac{1}{2}\times 10\times t^2$

$t=\frac{1}{\sqrt{5}}s$