Question

A block whose mass is 500g is fastened to a spring . The spring has spring constant 100N/m . The block is pulled to a distance of x=10 cm from its equilibrium position state of x=0 from rest at t=0. What is kinetic energy of block at x=5cm?

Answer 1

Step 1, Given Data

Mass of block, $m=500g=0.5kg$

Spring constant, $k=100N{m}^{-1}$

Amplitude of block,$A=10cm=0.1m$

Kinetic energy of block at $x=5cm=0.05m$

Step 1: Finding kinetic energy:

Spring will show simple harmonic motion so,

$K.E=\frac{1}{2}m{v}^2$

Relatipn of velocity in S.H.M is given as;

$v=w\sqrt{A^2-x^2}$

$K.E=\frac{1}{2}m{\left(w\sqrt{A^2-x^2}\right)}^2$

$m{w}^2=k$

$K.E.=\frac{1}{2}k(A^2-x^2)$

$K.E.=\frac{1}{2}\left(100\right)({0.1}^2-{0.05}^2)$

$K.E.=\frac{0.75}{2}=0.375J$

Hence the kinetic energy is 0.375 J