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Question

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


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Solution

Step1: Given data

  1. The spring constant of the spring is K=800N/m.
  2. The initial extension of the spring is x1=5cm=0.05m.
  3. The final extension of the spring is x2=15cm=0.15m.

Step2: Work done on a spring

  1. The work done by a spring is the total amount of energy to extend the spring from one position to another.
  2. The force exerted on a spring due to extension is proportional to the amount of extension, i.e, Fx, where, x is the amount of extension.
  3. The work done by a spring is defined by the form, U=12Kx22-x12, where, K is the spring constant, x1 is the initial, and x2 is the final extension of the spring.

Step3: Diagram

A spring of force constant 800 Nm has an extension class 11 physics CBSE

Step4: Finding the work done

As we know, work done in spring due to extension is U=12Kx22-x12.

So,

U=12Kx22-x12=12×800×0.152-0.052orU=400×0.0225-0.0025orU=8joule.

Therefore, find work done in extending the spring from 5cm to 15cm is 8joule.


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