Question

Given $K_1$ = 1500 N/m, $K_2$ = 500 N/m $m_1$ = 2 kg, $m_2$ = 1 kg, Find Work done in slowly pulling down $m_2$ by 8 cm from equilibrium position.

• A. 0.4 J
• B. 0.5 J
• C. 1 J
• D. 10 J

Answer 1

The correct option is C

1 J

The initial extension in the springs of force constant $k_1$ and $K_2$, at equilibrium position:let be, $x_{20}$ and $x_{10}$.Then $x_{20}=\frac{m_{2}g}{k_2},x_{10}=\frac{(m_1+m_2)g}{k_1}$

Potential energy stored in the springs in equilibrium position is $U_1=\frac{1}{2}{k}_1{x}_{10}^2+\frac{1}{2}{k}_2{x}_{20}^2$

Put values of $x_{10}$,$x_{20}$ from above we get $U_1=0.4J$

Let $△x_1$ and $△x_2$ be additional elongation's due to pulling $m_2$ by l = 8cm.Additional forces on $m_1$ are equal and in opposite direction $\Rightarrow k_1(x_1+△x_1)=m_{1}g+k_2(x_2+△x_2)$

$\Rightarrow k_1△x_1=k_2△x_2$ ------------------(i)

Also $△x_1+△x_2=l$ ------------------------(ii)

$△x_1,△x_2$ can be found from (i) and (ii) $w_g+w_p+w_s$ = 0 (where $w_p$ is the work done by the pulling force) ($△k$ = 0)

$\Rightarrow w_p=-w_s-w_g=(U_2-U_1)-(m_{1}g△x_1+m_{2}g△x_2)$

Where $U_2=\frac{1}{2}{k}_1(x_{10}+△x_1{)}^2+\frac{1}{2}{k}_2(x_{20}+△x_2{)}^2$

Putting the values $\Rightarrow w_p=1J$