Figure (8-E7) shows a spring fixed at the bottom end of an incline of inclination $37^{\circ}$. A small block of mass $2 \mathrm{~kg}$ starts slipping down the incline from a point $4.8 \mathrm{~m}$ away from the spring. The block compresses the spring by $20 \mathrm{~cm}$, stops momentarily and then rebounds through a distance of $1 \mathrm{~m}$ up the incline. Find (a) the friction coefficient between the plane and the block and (b) the spring constant of the spring. Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$.
By work energy principle
(i) for downward motion,
$0-0=(-m g \sin \theta \times s)-\mu R s-\frac{\frac{1}{2}}{k x^{2}}$
$80 \mu+0.02 \mathrm{R}=60$
(ii) for upward motion
$0-0=(-m g \sin \theta \times s)-\mu R s+\frac{1}{2} k x^{2}$
$-16 \mu+0.02 k=12 \ldots \ldots(2)$
Solving 1 and 2 we get,
$\mu=0.5$
and
$k=1000 \mathrm{~N} / \mathrm{m}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.