Question

A spring is fixed at the bottom end of an incline of inclination $37^{\circ}$. A small block is released from rest on an incline from a point $4.8 \,m$ away from the spring. The block compresses the spring by $20\, cm$, stops momentarily and then rebounds through a distance of $1 \,m$ up the incline. The spring constant of the spring in $N / m$ is______. Take $g=10 \,m / s ^{2}$.

Answer 1

(a) Just after release

(b) When stopped for the first time

(c) When stopped for the second time

$W_{\text {gravity }}+W_{\text {friction }}+W_{\text {spring }}=\Delta k E=\frac{1}{2} m(0-0)=0$
$\therefore 20 \times 5 \sin 37^{\circ}-\mu\left(20 \cos 37^{\circ}\right) 5-\frac{1}{2} k\left[(0.2)^{2}-0\right]=0 \ldots$ (i)
Applying work energy for motion from (b) to (c)
$-20 \times 1 \times \sin 37^{\circ}-\mu\left(20 \cos 37^{\circ}\right) \times 1-\frac{1}{2} k\left[0-(0,2)^{2}\right]=0 $...(ii)
Adding Eqs. (i) and (ii)
$\therefore-20(5-1) \times \frac{3}{5}-\mu\left(20 \times \frac{4}{5}\right)(5+1)=0 $
$\Rightarrow \mu=0.5$
Putting this value in Eq. (i), we get $k=1000\, N / m$