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)

$0.0375\,J, 0.125 \,J$

Answer 2

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, $U=\frac{1}{2}kx^{2}$ $= \frac{1}{2}\times 100\times\left(0.05\right)^{2} = 0.125 \,J$