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 dataThe spring constant of the spring is $K=$$800N/m$.The initial extension of the spring is ${x}_{1}=5cm=0.05m.$The final extension of the spring is ${x}_{2}=15cm=0.15m.$Step2: Work done on a springThe work done by a spring is the total amount of energy to extend the spring from one position to another.The force exerted on a spring due to extension is proportional to the amount of extension, i.e, $F\propto x$, where, x is the amount of extension. The work done by a spring is defined by the form, $U=\frac{1}{2}K\left({{x}_{2}}^{2}-{{x}_{1}}^{2}\right)$, where, K is the spring constant, ${x}_{1}$ is the initial, and ${x}_{2}$ is the final extension of the spring.Step3: DiagramStep4: Finding the work doneAs we know, work done in spring due to extension is $U=\frac{1}{2}K\left({{x}_{2}}^{2}-{{x}_{1}}^{2}\right)$.So, $U=\frac{1}{2}K\left({{x}_{2}}^{2}-{{x}_{1}}^{2}\right)=\frac{1}{2}×800×\left[{\left(0.15\right)}^{2}-{\left(0.05\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}orU=400×\left[0.0225-0.0025\right]\phantom{\rule{0ex}{0ex}}orU=8joule.$Therefore, find work done in extending the spring from $5cm$ to $15cm$ is $8joule.$

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