Question

Given $$k_1 = 1500 \ N/m,$$ $$k_2 =500 \ N/m,$$ $$m_1 = 2 \ kg$$ and $$m_2 = 1 \ kg$$. Find potential energy stored in equilibrium: (Take $$g = 10 \ m/s^2$$)

Answer 1

Correct option is B. 0.4 J
Soppose the elongation in first string be $$x_1$$ and in the second string be $$x_2$$

Now for second string, in equilibrium

$$K_2x_2=m_2g$$

or $$x_2= m_2g/k_2$$ (1)

For first string in equilibrium

$$K_1x_1=(m_1+m_2)g$$

or $$x_1= (m_2+m_1)g/k_1$$ (2)

Now potential energy stored will be

$$U=\frac{K_1x_1^2}{2}$$ + $$\frac{K_2x_2^2}{2}$$

Putting the values of $$k_1$$,$$k_2$$,$$x_1$$ and $$x_2$$

We get U= 0.4J