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Question

It is given that k1=1500 N/m, k2=500 N/m, m1=2 kg and m2=1 kg. then, what is the total potential energy stored in the springs in equilibrium in Joule? (g=10 m/s2)
4286.png

A
0.3
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B
0.2
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C
0.4
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D
0.6
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Solution

The correct option is B 0.4
Let the extensions of the springs with spring constants k1 and k2 be x1 and x2 respectively.

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

For equilibrium,
k2x2=m2g
k1x1=m1g+k2x2

Putting the values in the above equations, we get x2=0.02m and x1=0.02m

So the potential energy stored in the springs will be

12k1x21+12k2x22=12(1500+500)(0.02)2=0.4 J

367969_4286_ans.png

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