Question

Both the blocks are resting on a horizontal floor and the pulley is held such that the string remains just taut. At the moment $t=0$, a force $F=20t\text{N}$ starts acting on the pulley along vertically upward direction as shown in figure. Calculate the velocity of A as B loses contact with the floor.
[Take $g=10{\text{m/s}}^2$]

Answer 1

Let $T$ be the tension in the string.
Then, $2T=F=20t$ or $T=10t\text{N}$
Let the block A loses its contact with the floor at time $t=t_1$.This happens when the tension in the string becomes equal to the weight of A.
i.e $T=m_{A}g$ or $10t_1=1\times 10$
or $t_1=1\text{s}......\left(i\right)$

Similarly for block B, we have
$10t_2=2\times 10$ or $t_2=1\text{s}......\left(ii\right)$
i.e, the block B loses contact after $2\text{s}$.

For block A, at time $t$ such that $t\ge t_1$, let $a$ be its accelaration in upward direction.
Then, $10t-1\times 10=1\times a=\frac{dv}{dt}$
or $dv=10(t-1)dt.......\left(iii\right)$

Integrating this expression, we get
${\int }_0^{v}dv=10{\int }_1^t(t-1)dt$
or $v=5t^2-10t+5$
Substituting $t=t_2=2\text{s}$
or $v=20-20+5=5\text{m/s}.....\left(v\right)$