Identical springs of steel and copper are equ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The work done in stretching a spring is given by $W = \frac{1}{2} k (\Delta l)^2$,where $k$ is the spring constant and $\Delta l$ is the extension

Vedclass · Accepted Answer

Copper spring

Vedclass · Answer

(B) The work done in stretching a spring is given by $W = \frac{1}{2} k (\Delta l)^2$,where $k$ is the spring constant and $\Delta l$ is the extension.Since the springs are identical,they have the same dimensions ($l$ and $A$). The spring constant $k$ is given by $k = \frac{YA}{l}$,where $Y$ is Young's modulus.Therefore,$W = \frac{1}{2} \left( \frac{YA}{l} ight) (\Delta l)^2$.Since $A, l,$ and $\Delta l$ are constant for both springs,we have $W \propto Y$.Because the Young's modulus of steel $(Y_{	ext{steel}})$ is greater than that of copper $(Y_{	ext{copper}})$,it follows that $W_{	ext{steel}} > W_{	ext{copper}}$.Wait,let us re-evaluate: If the springs are stretched by the same *force* $F$,then $\Delta l = \frac{Fl}{AY} \propto \frac{1}{Y}$. The work done $W = \frac{1}{2} F \Delta l \propto \Delta l \propto \frac{1}{Y}$. Since $Y_{	ext{steel}} > Y_{	ext{copper}}$,then $W_{	ext{steel}} Thus,more work is done on the copper spring.

Identical springs of steel and copper are equally stretched. On which more work will have to be done?

Similar Questions

The elastic potential energy stored in a steel wire of length $20 \, m$ stretched by $2 \, cm$ is $80 \, J$. The cross-sectional area of the wire is $......... \, mm^2$ (Given, $Y = 2.0 \times 10^{11} \, N/m^2$).

On stretching a wire,the elastic energy stored per unit volume is

When a $8 \,m$ long wire is stretched by a load of $10 \,kg-wt$, it is elongated by $1.5 \,mm$. The energy stored in the wire in this process is $\left(g=10 \,ms^{-2}\right)$ (in $\,J$)

$A$ rubber pipe of density $1.5 \times 10^3 \, kg/m^3$ and Young's modulus $5 \times 10^6 \, N/m^2$ is suspended from the roof. The length of the pipe is $8 \, m$. What will be the change in length due to its own weight?

$A$ wire fixed at the upper end stretches by length $l$ by applying a force $F$. The work done in stretching is

Vedclass Test Series

Exam Paper Generator

Online Exam Module