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Question

A spring stores 1J of energy for compression of 1mm. The additional work to be done to compress it further by 1mm is


A

1J

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B

2J

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C

3J

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D

4J

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E

0.5J

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Solution

The correct option is C

3J


Step 1: Given data

Energy, W1=1J

Compression x1=1mm

Compression x2=2mm

Step 2: Find the additional work done

The work done is given by,

W1=(1/2)kx121=(1/2)k(1×10-3)2k=2×106

After additional compression, the work done is

W2=(1/2)kx22W2=(1/2)(2×106)(2×10-3)2W2=4J

The additional work done is

=W2W2=4-1=3J

Hence, option C is correct.


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