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Question

# A spring of spring constant 50 N/m initially at its natural length is compressed through 1 cm. Find the work done by the spring-force on the external agent, compressing the spring.

A
2.5 mJ
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B
2.5 mJ
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C
3.5 mJ
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D
3.5 mJ
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Solution

## The correct option is A −2.5 mJThe spring force is the restoring force given by F=−kx and it will act in a direction opposite to the displacement of the point on the spring. Therefore, work done by spring force, Wspring=∫Fspringdx Wspring=∫x0−kxdx=−12kx2 ⇒W=−12×(50 N/m)×(0.01 m)2 Or, W=−2.5×10−3 J=−2.5 mJ The force will be opposite to the displacement of the point of application, hence it is evident that work done by spring-force will be negative. Alternative solution:––––––––––––––––––––––– Spring force is a conservative force, hence we can write Wspring=−ΔU ′U′ represents the spring potential energy When spring was at its natural length, Ui=0 After compressing the spring through distance x, Uf=12kx2 ⇒Wspring=−(Uf−Ui)=−Uf Or, Wspring=−12kx2 Or, Wspring=−12×(50 N/m)×(0.01 m)2 ∴Wspring=−2.5×10−3 J=−2.5 mJ

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