In the arrangement shown in the figure, work done by the string on the block of mass \(0.36 \mathrm{~kg}\)…
- \(8 \mathrm{~J}\)
- \(4 \mathrm{~J}\)
- \(12 \mathrm{~J}\)
- \(2 \mathrm{~J}\)
Solution
Given, mass of block \(1, m_1=0.36 \mathrm{~kg}\) mass of block \(2, m_2=0.72 \mathrm{~kg}\) and acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2\) Now, Dynamic equation of block \(m_1\left(m_2 > m_1\right)\). \(\therefore \quad T-m_1 g=m_1 a\)...(i) Dynamic equation of block \(m_2\left(m_2 > m_1\right)\). \(\therefore \quad m_2 g-T=m_2 a \quad \ldots (ii)\) Thus, from adding Eqs. (i) and (ii), we get \(\therefore \quad T-m_1 g+m_2 g-T=m_1 a+m_2 a\) \(m_2 g-m_1 g=a\left(m_1+m_2\right)\) 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, \(a=\frac{(0.72-0.36)}{1.08} \times 10\) \(a=\frac{3.6}{1.08} \mathrm{~m} / \mathrm{s}^2 \quad \ldots (iii)\) Now, the tension force in the string is, for block 1st, \(m_1=0.36\) \(\begin{aligned} & T-m_1 g=m_1 a \\ & \frac{T-m_1 g}{m_1}=a \\ & \frac{T-3.6}{0.36}=\frac{3.6}{1.08} \\ & T=3.6+0.36 \times \frac{3.6}{1.08} \\ & T=4.8 \quad \ldots (iv) \end{aligned}\) Distance travelled by mass \(m_1\) block, By displacement-time relation, Displacement, \(s=u t+\frac{1}{2} \mathrm{at}^2\) \([\because u=0]\) [by kinetic equation for uniformly motion] \(s=\frac{1}{2} \times \frac{3.6}{1.08} \times(1)^2 \quad[\therefore t=1\) second given] \(s=\frac{1}{2} \times \frac{3.6}{1.08}\)...(v) Now, work done by the string on the block of mass \(\left(m_1\right)\) is \(0.36 \mathrm{~kg}\). \(\therefore\) 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 \(\begin{aligned} & W=4.8 \times \frac{1}{2} \times \frac{3.6}{1.08} \\ & W=\frac{16}{2} \Rightarrow W=8 \mathrm{~J} \end{aligned}\)
Asked in: AP EAMCET 2019 (22 Apr Shift 1)