Question

# A wire suspended vertically from one of Its ends is stretched by attaching a weight of $\text{200}$N to the Lower End. The weight stretches the wire by $1$mm. Then the elastic energy stored in the wire is ?

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Solution

## Step 1: Given parametersForce, F = $\text{200}$N.Stretching of wire, x = $1$mm.Step 2: Formula usedElastic energy$=\frac{1}{2}×\mathrm{F}×\mathrm{x}$ where F is the force and x is the extension produced.Step 3: Calculating the elastic energy stored in the wire.By substituting the given values, we get $=\frac{1}{2}×200×1×{10}^{-3}\phantom{\rule{0ex}{0ex}}=0.1\mathrm{J}$Therefore, the elastic energy will be $0.1\mathrm{J}$.

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