Question

In the arrangement shown in the figure, work done by the string on the block of mass $0.36kg$ 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, $g$ $.=10m{s}^{-2}$)

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

Answer 1

Answer: (A)

According to the question,

Given, mass of block $1,m_1=0.36kg$
mass of block $2,m_2=0.72kg$
and acceleration due to gravity, $g=10m/s^2$
Now, Dynamic equation of block $m_1\left(m_2>m_1\right)$.
$\therefore T-m_{1}g=m_{1}a$ ..........(i)
$\dots$ (i) Dynamic equation of block $m_2\left(m_2>m_1\right)$.
$\therefore m_{2}g-T=m_{2}a\dots (ii)$
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, $s=ut+\frac{1}{2}a{t}^2[\because u=0]$
[by kinetic equation for uniformly motion]
$s=\frac{1}{2}\times \frac{3.6}{1.08}\times (1{)}^2[\therefore t=1\text{ second given }]$
$s=\frac{1}{2}\times \frac{3.6}{1.08}\dots (v)$
Now, work done by the string on the block of mass $\left(m_1\right)$ is $0.36kg$.
$\therefore$ 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
$W=4.8\times \frac{1}{2}\times \frac{3.6}{1.08}$
$W=\frac{16}{2}\Rightarrow W=8J$