Question

A piece of $1kg$ mass is hanged from a spring. The spring constant is $50N/m.$ The piece is stretched from the equilibrium position $x=0,t=0$ on a frictional surface to a distance $x=10cm.$ The piece is at a distance $5cm$ from its mean position. Calculate its kinetic, potential and total energy.

Answer 1

Given $m=1kg;k=50N{m}^{-1}$

$a=10cm=0.1m$

$y=5cm=0.05mthenK=?,U=?E_{total}=?$

Angular speed ${\omega }_0=\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{1}}=5\sqrt{2}$

$\therefore$ When $y=0.05m$ kinetic energy

$K=\frac{1}{2}m{\omega }_0^2(a^2-y^2)$

or $K=\frac{1}{2}\times 1\times 50\left[\right(0.1{)}^2-\left(0.05{)}^2\right]$

$=25(0.1+0.05)(0.1-0.05)$

$=25\times 0.15\times 0.05$

$=0.1875J$

or $K=0.19J$

Potential energy $U=\frac{1}{2}m{\omega }_0^2{y}^2$

$=\frac{1}{2}\times 1\times 50\times 0.05\times 0.05$

$=25\times 0.05\times 0.05$

$=25\times 0.0025$

or $U=0.0625J$

Total energy , $E_{total}=\frac{1}{2}m{\omega }_0^2{a}^2$

$=\frac{1}{2}\times 1\times 50\times 0.1\times 0.1=25\times 0.01$

$E_{total}=0.25J$