Question

A metal wire length $L$, cross sectional area $A$ and Young's modulus $Y$ is stretched by a variable force $F$. $F$ is varying in such a way that $F$ is always slightly greater than the elastic forces or resistance in the wire. When the elongation in the wire is $l$, up to this instant

• A. The work done by $F$ is $\frac{YA{l}^2}{2L}$
• B. The work done by $F$ is $\frac{YA{l}^2}{L}$
• C. the elastic potential energy stored in wire is $\frac{YA{l}^2}{2L}$
• D. No energy is lost during elongation

Answer 1

Answer: (A), (C), (D)

The correct options are
A The work done by $F$ is $\frac{YA{l}^2}{2L}$
C the elastic potential energy stored in wire is $\frac{YA{l}^2}{2L}$
D No energy is lost during elongation

In this question, the applied force is not constant.
Let us consider at any instant the wire elongates by $x$. Then work done by $F$ to elongate it more by $dx$ is
$dW=Fdx=\frac{YAx}{L}dx[x<<L]$
$W={\int }_0^l\frac{YAxdx}{L}=\frac{YA{l}^2}{2L}$
As internal restoring force and applied force are almost same (equal and opposite), magnitude of work done by these forces would be equal.