Both the blocks are resting on a horizontal floor and the pullcy is held such that string remains just taut. At moment $t=0$, a force $F=20 t \mathrm{~N}$ starts acting on the pullcy along vertically upward direction as shown in figurc. Calculate:
(a) velocity of $A$ when $B$ loses contact with the floor.
(b) hcight raised by the pulley upto that instant. $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
Solution
(a) Let 'I' be the tension in the string. Then, $27=20$ tor $T=10 t \mathrm{~N}$ Let the block $A$ loses its contact with the floor at time $t=t_{1}$. This happens when the tension in string becomes equal to the weight of $A$. Thus, $T=m g$ or $10 t_{1}=1 \times 10$ or
$t_{1}=1 \mathrm{~s} \quad \ldots(1)$
Similarly, for block $B$, we have $10 t_{2}=2 \times 10 \quad$ or $\quad t_{2}=2 \mathrm{~s} \quad \ldots$ (ii)
\mathrm{\{} i e . , ~ t h e ~ b l o c k ~ $B$ loses contact alter $2 \mathrm{~s}$. For block $A$, at time $t$ such that $t \geq t$ let $a$ be its acceleration in upward direction. Then,
$10 t-1 \times 10=1 \times a=(d v / d t) \quad$ or $\quad d v=10(t-1) d t$
Integrating this cxpression, we get
$\int_{0}^{\mathrm{v}} d v-10 \int_{1}^{t}(t-1) d t$
or $v=5 t^{2}-10 t+5 \quad \ldots(1 \mathrm{v})$
Substituting $t=t_{2}=2 \mathrm{~s} \quad$ or $\quad v=20-20+5=5 \mathrm{~m} / \mathrm{s} \ldots(\mathrm{v})$
(vi) (b) From Eq. (iv), $d y=\left(5 t^{2}-10 t+5\right) d t$
Where $y$ is the vertical displacemcnt of block $A$ at timc $t\left(\geq t_{1}\right)$. Intcgrating, we have
$\int_{v=0}^{<=h} d y-\int_{t=1}^{t=2}\left(5 t^{3}-10 t 15\right) d i \quad \Rightarrow h-5\left[\frac{t^{3}}{3}\right]_{1}^{2} 10\left[\frac{t^{2}}{2}\right]_{1}^{3} 15\lfloor t]_{1}^{3}-\frac{5}{3} \mathrm{~m}$
$\therefore$ lleight raised by pulley upto that instant $-\frac{h}{2}-\frac{5}{6} \mathrm{~m}$