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Given $K_{1} = 1500 N/m and K_{2} = 500 N/m .$ $m_{1} = 2 kg and m_{2} = 1 kg (g = 10 {m/sec}^{2})$ . Then find the total potential energy stored in the springs in equilibrium

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0.3 J

0.2 J

0.4 J

0.6 J

answer is C.

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Detailed Solution

m_{2} g = k_{2} x_{2}, x_{2} = \frac{1}{50}

$k_{1} x_{1} = (m_{1} + m_{2}) g$

$x_{1} = \frac{30}{1500} = \frac{1}{50}$

$U = \frac{1}{2} k_{1} x_{1}^{2} + \frac{1}{2} k_{2} x_{2}^{2}$

$U = 0.3 + 0.1$

$U = 0.4 J$

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