Given $K_{1}$ = 1500 N/m, $K_{2}$ = 500 N/m $m_{1}$ = 2 kg, $m_{2}$ = 1 kg, Find Potential energy stored in the springs in equilibrium.

0.4 J

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0.5 J

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1 J

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10J

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Solution

The correct option is A
0.4 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 $U_{1} = 0.4 J$

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Given K1 = 1500 N/m, K2 = 500 N/m m1 = 2 kg, m2 = 1 kg, Find Potential energy stored in the springs in equilibrium.

Given $K_{1}$ = 1500 N/m, $K_{2}$ = 500 N/m $m_{1}$ = 2 kg, $m_{2}$ = 1 kg, Find Potential energy stored in the springs in equilibrium.