Two similar springs P and Q have spring constants KP and KQ, such that KP > KQ. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs WP and WQ are related as, in case (a) and case (b), respectively:

  • Option 1)

    WP > WQ; WQ > WP

  • Option 2)

    WP < WQ; WQ < WP

  • Option 3)

    WP= WQ; WP > WQ

  • Option 4)

    WP = WQ; WP = WQ

 

Answers (1)

As we discussed in concept

Work done by restoring force -

W= -frac{1}{2}: kx^{2}

- wherein

Restoring force f=-kx (spring force)

 

 w=\frac{1}{2}kx^{2}

 

case (a) if extension (x) is same

w_{p}=\frac{1}{2}k_{p}x^{2}           so w_{p}> w_{Q}

w_{Q}=\frac{1}{2}k_{Q}x^{2}

case (b) if spring force (F) is same (force of elongation)

x_{1}=\frac{F}{k_{p}} \: and\: x_{2}=\frac{F}{k_{Q}}

w_{p}=\frac{1}{2}k_{p}x_{1}^{2}

=\frac{1}{2}\frac{F^{2}}{k_{p}}

w_{Q}=\frac{1}{2}k_{Q}x_{2}^{2}

=\frac{1}{2}\frac{F^{2}}{k_{Q}}

\therefore w_{p}< w_{Q}

 


Option 1)

WP > WQ; WQ > WP

Option 2)

WP < WQ; WQ < WP

Option 3)

WP= WQ; WP > WQ

Option 4)

WP = WQ; WP = WQ

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