Question

A block whose mass is $1$kg is fastened to a spring. The spring has a spring constant of $100N{m}^{-1}$. The block is pulled to a distance $x=10$cm from its equilibrium position at $x=0$ on a frictionless surface from rest at $t=0$. The kinetic energy and potential energy of the block when it is $5$cm away from the mean position is

• A. $0.0375J,0.125J$
• B. $0.125J,0.375J$
• C. $0.125J,0.125J$
• D. $0.375J,0.375J$

Answer 1

Answer: (C)

$0.125J,0.125J$

Mass$=1kg$

Spring constant$=10N/m$

$x=10cmx=0t=0$

Kinetic energy and potential energy of the block$=5cm$ away from the mean position

Angular frequency$\omega =\sqrt{k/m}\omega =\sqrt{50/1}=7.07rad/s$

Its displacement at any time t is then given by$x\left(t\right)=0.1\cos \left(7.07t\right)$

$P.E=\frac{1}{2}k{x}^2=\frac{1}{2}\times 100\times (0.1{)}^2=\frac{1}{2}\times 100\times \frac{1}{1000}=0.125J$