Question

In the arrangement shown in the figure, work done by the string on the block of mass $0.36\, kg$ during the first second after the blocks are released from state of rest is (Ignore friction and mass of the string.)
(Acceleration due to gravity, $g$ $.=10 ms ^{-2}$)

• A. $8\, J$
• B. $4\, J$
• C. $12 \,J$
• D. $2 \,J$

Answer 1

Answer: (A)

According to the question,

Given, mass of block $1, m_{1}=0.36\, kg$
mass of block $2, m_{2}=0.72 \,kg$
and acceleration due to gravity, $g=10 \,m / s ^{2}$
Now, Dynamic equation of block $m_{1}\left(m_{2}>m_{1}\right)$.
$\therefore T-m_{1} g=m_{1} a$ ..........(i)
$\ldots$ (i) Dynamic equation of block $m_{2}\left(m_{2}>m_{1}\right)$.
$\therefore \quad m_{2} g-T=m_{2} a \ldots(ii)$
Thus, from adding Eqs. (i) and (ii), we get
$\therefore T-m_{1} g+m_{2} g-T=m_{1} a+m_{2} a$
$m_{2} g-m_{1} g=a\left(m_{1}+m_{2}\right)$
or $ a=\frac{\left(m_{2}-m_{1}\right) g}{\left(m_{1}+m_{2}\right)}$
Hence, acceleration, $a=\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right) g$
$\therefore $ Acceleration, $a=\frac{(0.72-0.36)}{1.08} \times 10$
$a=\frac{36}{1.08} m / s ^{2}$
Now, the tension force in the string is, for block lst, $m_{1}=0.36$
$T-m_{1} g=m_{1} a$
$\frac{T-m_{1} g}{m_{1}}=a$
$\frac{T-3.6}{0.36}=\frac{3.6}{1.08} $
$T=3.6+0.36 \times \frac{3.6}{1.08}$
$T=4.8$ ......(IV)
Distance travelled by mass $m_{1}$ block,
By displacement-time relation,
Displacement, $s=u t+\frac{1}{2} at ^{2} [\because u=0]$
[by kinetic equation for uniformly motion]
$s=\frac{1}{2} \times \frac{3.6}{1.08} \times(1)^{2} [\therefore t=1 \text { second given }] $
$s=\frac{1}{2} \times \frac{3.6}{1.08} \ldots( v )$
Now, work done by the string on the block of mass $\left(m_{1}\right)$ is $0.36\, kg$.
$\therefore $ Work done, $W=$ Tension force $(T) \times$ displacement $(s)$
$W=T \times \frac{1}{2} a t^{2}$
From Eqs. (iv) and (v), we get
$W=4.8 \times \frac{1}{2} \times \frac{3.6}{1.08} $
$W=\frac{16}{2} \Rightarrow W=8 \,J$