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 $4.8m$ away from the spring. The block compresses the spring by $20cm$, stops momentarily and then rebounds through a distance of $1m$ up the incline. The spring constant of the spring in $N/m$ is______. Take $g=10m/s^2$.

Answer 1

Answer: 1000

(a) Just after release

(b) When stopped for the first time

(c) When stopped for the second time

$W_{\text{gravity }}+W_{\text{friction }}+W_{\text{spring }}=\Delta kE=\frac{1}{2}m(0-0)=0$
$\therefore 20\times 5\sin 37^{\circ }-\mu \left(20\cos 37^{\circ }\right)5-\frac{1}{2}k\left[(0.2{)}^2-0\right]=0\dots$ (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 $k=1000N/m$