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. 0.375 J, 0.125 J
• B. 0.125 J, 0.375 J
• C. 0.125 J, 0.125 J
• D. 0.375 J, 0.375 J

Answer 1

Answer: (A)

0.375 J, 0.125 J

Answer 2

Here, $m = 1kg,k = 100Nm^{- 1}$

$A = 10cm = 0.1m$

The block executes SHM, its angular frequency is given by

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100Nm^{- 1}}{1kg}} = 10rads^{- 1}$

Velocity of the block at x = 5 cm $(( = 0.05m)$ is

$v = \omega\sqrt{A^{2} - x^{2}} = 10\sqrt{(0.1)^{2} - (0.05)^{2}}$

$= 10\sqrt{7.5 \times 10^{- 3}}ms^{- 1}$

Kinetic energy of the block,

$K = \frac{1}{2}mv^{2} = \frac{1}{2} \times 1 \times 0.75 = 0.375J$

Potential energy of the block,

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