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Problem Solving

It is given t...

Question

It is given that $k_{1} = 1500$ N/m, $k_{2} = 500$ N/m, $m_{1} = 2$ kg and $m_{2} = 1$ kg. then, what is the total potential energy stored in the springs in equilibrium in Joule? $(g = 10 m / s^{2})$

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Solution

The correct option is B $0.4$
Let the extensions of the springs with spring constants $k_{1}$ and $k_{2}$ be $x_{1}$ and $x_{2}$ respectively.

The forces acting on the body are as shown in the figure.

For equilibrium,
$k_{2} x_{2} = m_{2} g$
$k_{1} x_{1} = m_{1} g + k_{2} x_{2}$

Putting the values in the above equations, we get $x_{2} = 0.02 m$ and $x_{1} = 0.02 m$

So the potential energy stored in the springs will be

$\frac{1}{2} k_{1} x_{1}^{2} + \frac{1}{2} k_{2} x_{2}^{2} = \frac{1}{2} (1500 + 500) (0.02)^{2} = 0.4 J$

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Similar questions

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.

k_{1}

k_{2}

m_{1}

m_{2}

m_{2}

k_{1} = 1500 N m^{- 1}, k_{2} = 500 N m^{- 1}, m_{1} = 2 k g, m_{2} = 1 k g

m_{2}

8 c m

m

l

k_{1}

k_{2}

rad/s

Take l = 1 m, b = \frac{1}{2 \sqrt{2}} m, k_{1} = 16 N/m, k_{2} = 1 N/m, m = \frac{1}{64} kg

m_{2} i n {m/s}^{2}

m_{1} = 2 k g, m_{2} = 6 k g, F = 93 N, g = 10 m / s^{2},

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