Question

A spring connected between two blocks of mass 2 kg and 4 kg is compressed by 20 cm & released. After time $t_0$, the spring gains natural length. Find the work done by the spring on the 2kg block between t = 0 & $t=t_0$.

• A. $\frac{40}{3}$ J
• B. 20 J
• C. 40 J
• D. $\frac{20}{3}$ J

Answer 1

The correct option is A $\frac{40}{3}$ J


$Initialmomentum{p}_i=0$


$Finalmomentum{p}_f=-2v_1+4v_2$
Using conservation of momentum,
$-2v_1+4v_2=0\Rightarrow v_1=2v_2$

By conservation of energy,

$\frac{1}{2}k{x}^2=\frac{1}{2}{m}_1{v_1}^2+\frac{1}{2}{m}_2{v_2}^{2}wherexistheinitialcompressionofthespring\Rightarrow \frac{1}{2}k{(0.2)}^2=\frac{1}{2}{m}_1{v_1}^2+\frac{1}{2}{m}_2\left(\frac{{v_1}^2}{4}\right)(\because v_2=\frac{v_1}{2})\Rightarrow \frac{1}{2}\times 1000\times {(0.2)}^2=\frac{1}{2}\times 2\times {v_1}^2+\frac{1}{2}\times 4\times \left(\frac{{v_1}^2}{4}\right)\Rightarrow 20=\frac{3}{2}{v_1}^2\Rightarrow \frac{1}{2}\times 2\times {v_1}^2=\frac{40}{3}=\frac{1}{2}{m}_1{v_1}^2\therefore Workdoneon2kgblock=Changeinkineticenergy=\frac{40}{3}-0=\frac{40}{3}J$