Question

A wire 2 m in length suspended vertically stretches by 10 mm when the mass of 10 kg is attached to the lower end. The elastic potential energy gain by the wire is? (take g = 10 $m/s^2$)

• A. 0.5 J
• B. 5 J
• C. 50 J
• D. 500 J

Answer 1

The correct option is A 0.5 J

Formulae:

spring force by wire = weight of mass attached

$k\Delta x=mg$

$k\left(0.01\right)=10\times 9.8$

$k=9800N/m$

elastic potential energy,

$U=\left(0.5\right)k\Delta x^2$

$=\left(0.5\right)\times 9800\times (0.01{)}^2$

$\therefore U=0.49\approx 0.5J$