Question

A block whose mass is $1 \,kg$ is fastened to a spring. The spring has a spring constant of $100 \,N m^{-1}$. 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. $0.0375\,J, 0.125 \,J$
• B. $0.125\,J, 0.375 \,J$
• C. $0.125\,J, 0.125 \,J$
• D. $0.0375\,J, 0.375 \,J$

Answer 1

Answer: (A)

Here, $m = 1 \,kg, k = 100 \,N \,m ^{-1}$
$A = 10 \,cm = 0.1\, m$
The block executes $SHM$, its angular frequency is given by
$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100 N m^{-1}}{1 kg}} = 10 \,rad s^{-1} $
Velocity of the block at $x =5 \,cm = 0 .05 \,m$ is
$ v = \omega\sqrt{A^{2} -x^{2}} $
$ =10\sqrt{\left(0.1\right)^{2}-\left(0.05\right)^{2}}$
$= 10\sqrt{7.5\times10^{-3}}m s^{-1} $
Kinetic energy of the block,
$K =\frac{1}{2}mv^{2} $
$ = \frac{1}{2}\times 1\times 0.75 $
$= 0.0375 \,J $
Potential energy of the block,
$U=\frac{1}{2}kx^{2} $
$ = \frac{1}{2}\times 100\times\left(0.05\right)^{2} = 0.125 \,J$