Question

Two similar springs P and Q have spring constants $k_{P}\ and\ k_{Q}\ such\ that\ k_{P} \ \textgreater \ k_{Q}$. They are stretched, first by the same amount (case a,) then by the same force (case b). The work done by the springs $W_{P}\ and\ W_{Q}$ are related as, in case (a) and case (b), respectively(a) $W_{P}=W_{Q};\ W_{P}=W_{Q}$(b) $W_{P}\ \textgreater \ W_{Q};\ W_{Q}\ \textgreater \ W_{P}$(c) $W_{P}\ \textless \ W_{Q};\ W_{Q}\ \textless \ W_{P}$(d) $W_{P}=W_{Q};\ W_{P}\ \textgreater \ W_{Q}$

Answer 1

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