Question

In the given figure a block of mass $m$ is tied on a wedge by an ideal string as shown in the figure. String is parallel towards to the inclined surface of wedge. All the surfaces involved are smooth. Wedge is being moved towards wight with a time varying velocity $v=\frac{t^2}{2}\left(m/s\right)$ where $t$ is in sec. At what time block will just break the contact with wedge ( use $g=10m/s$ )

• A. $10$ sec
• B. $5$ sec
• C. $2$ sec
• D. $4$ sec

Answer 1

The correct option is A $10$ sec

The block will lift once the pseudo force acting on the wedge becomes equal to the weight of the block.

As the wedge is moving with the velocity of$\frac{t^2}{2}m/s$, its acceleration will become

$\frac{d}{dt}\left(\frac{t^2}{2}\right)=tm/s^2$.

The component of weight along the surface is equal to $mg\cos {45}^{\circ }$.

The pseudo force acting opposite to it will be $mt\sin {45}^{\circ }$

Both the forces will be equal at the time when the block lifts off the wedge.

$mt\sin {45}^{\circ }=mg\cos {45}^{\circ }$

$t=g$

So at time $t=10\sec$ the block will lift from the wedge.