Two blocks A and B of masses $1 \mathrm{kg}$ and $2 \mathrm{kg}$ respectively are connected by a string, passing over a light frictionless pulley. Both the blocks are resting on a horizontal floor and the pulley is held such that string remains just taut. At moment t=0, a force F=20t newton starts acting on the pulley along vertically upward direction as shown in figure. Calculate height raised by the pulley up to that instant.

(Take, $g=10 \mathrm{m}/{\mathrm{s}}^2$ )

Let T be the tension in the string. Then,

$$\begin{gathered} 2T=20t \\ T=10t\text{ newton } \end{gathered}$$

$$\begin{gathered} \text{ Let the block }A\text{ loses its contact with the floor at time }t=t_1\text{. This happens when the tension } \\ \text{ in string becomes equal to the weight of }A\text{. Thus, } \\ T=mg \\ 10t_1=1\times 10 \\ {t}_1=1 \mathrm{s} \end{gathered}$$                                        …………………………….(i)

$$\begin{gathered} \text{ or } \\ \text{ Similarly, for block }B\text{, we have } \\ \text{ or }10t_2=2\times 10 \end{gathered}$$                             …………………………(ii)

$$\begin{gathered} \text{ i.e. the block }B\text{ loses contact at }2 \mathrm{s}.\text{ For block }A,\text{ at time }t\text{ such that }t\ge t_1\text{ let }a\text{ be its } \\ \text{ acceleration in upward direction. Then, } \\ 10t-1\times 10=1\times a=(dv/dt) \\ \text{ or } \end{gathered}$$

$\text{ Integrating this expression, we get }$

$\text{ Integrating this expression, we get}$                            ……………..(iii)

Substituting $t=t_2=2 \mathrm{s}$

$v=5t^2-10t+5$                                                                       

$v=20-20+5=5 \mathrm{m}/\mathrm{s}$                              …………………(iv)

From Eq. (iv), $dy=\left(5t^2-10t+5\right)dt$               ……………….(v)

where, y is the vertical displacement of block A at time $t\left(\ge t_1\right)$. Integrating, we have

${\int }_{y=0}^{y=h}dy={\int }_{t=1}^{t=2}\left(5t^2-10t+5\right)dt$

$h=5{\left[\frac{t^3}{3}\right]}_1^2-10{\left[\frac{t^2}{2}\right]}_1^2+5[t{]}_1^2=\frac{5}{3}m$

$\therefore$ Height raised by pulley upto that instant $=\frac{h}{2} =\frac{5}{6} \mathrm{m}$