First slide

Elastic behaviour of solids

Question

A mass m is hanging from a wire of cross-sectional area A and length L. Y is Young's modulus of wire. An external force F is applied on the wire which is then slowly further pulled down by $∆$ x from its equilibrium position. Find the work done by the force F the wire exerts on the mass.

Moderate

A

$mgΔx + \frac{YA}{2 L} (Δx)^{2}$
B

$mgΔx + \frac{YA (ΔX)^{2}}{L}$
C

$mg \frac{Δx}{2} + \frac{YA (Δx)^{2}}{L}$
D

$mg \frac{Δx}{2} + \frac{YA (Δx)^{2}}{2 L}$

Solution

Work done by applied force and gravity will be
$\begin{array}{l} W = \int_{0}^{Δx} K (x + Δl) dx = K {[\frac{x^{2}}{2} + xΔl]}_{0}^{Δx} \\ = \frac{YA}{L} [\frac{(Δx)^{2}}{2} + (Δx) (Δl)] \end{array}$
It will be work done by the force the wire exerts on the mass.
$= | - W | = W = \frac{YA}{L} [\frac{(Δx)^{2}}{2} + ΔxΔl]$
$\begin{array}{l} But Δl = \frac{mgl}{AY} \\ So, W = \frac{YA}{L} Δx [\frac{Δx}{2} + \frac{mgL}{AY}] \\ = \frac{YA}{L} \frac{(Δx)^{2}}{2} + mgΔx \end{array}$

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