Question:

Two springs have their force constant as $k_1$ and $k_2 (k_1 > k_2)$. when they are stretched by the same force

Updated On: Mar 22, 2024
  • no work is done in case of both the springs
  • equal work is done in case of both the springs
  • more work is done in case of second spring
  • more work is done in case of first spring
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The Correct Option is C

Solution and Explanation

Since force in both springs is same: $k _{1} \,x _{1}= k _{2} \,x _{2}$ If $k _{1}> k _{2}$ then $x _{1}< x _{2}$ Work $=0.5 \,kx ^{2}$ Hence, work by second spring is greater than first
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Concepts Used:

Hooke’s Law

Hooke’s Law states that for small deformities, the stress and strain are proportional to each other. Thus,

Stress ∝ Strain

Stress = k × Strain … where k is the Modulus of Elasticity.

When a limited amount of Force or deformation is involved then concept of Hooke’s Law is only applicable . If we consider the fact, then we can deviate from Hooke's Law. This is because of their extreme Elastic limits.