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  4. A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is:

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Q. A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is:

JamiaJamia 2005

Solution:

The work is stored as the P.E. of the body and is given by, $ U=\int_{{{x}_{1}}}^{{{x}_{2}}}{{{F}_{external}}}dx $ Or $ U=\int_{{{x}_{1}}}^{{{x}_{2}}}{kx}\,dx $ $ =\frac{1}{2}k(x_{2}^{2}-x_{1}^{2}) $ $ =\frac{800}{2}[{{(0.15)}^{2}}-{{(0.05)}^{2}}] $ $ [k=800(given)] $ $ =400[0.2\times 0.1] $ $ =8\,joule $