Question

A spring is fixed at the bottom end of an incline of inclination ${37}^{\circ }$. A small block of mass $2\text{kg}$ starts slipping down the incline from a point $4.8\text{m}$ away (along the inclined plane) from the spring. The block compresses the spring by $20\text{cm}$, stops momentarily and then rebounds through a distance of $1\text{m}$ up the incline. Then:

• A. Friction co-efficient between the plane and block is $0.7$
• B. Friction co-efficient between the plane and block is $0.5$
• C. Spring constant is $1000\text{N/m}$
• D. Spring constant is $800\text{N/m}$

Answer 1

Answer: (B), (C)

The correct options are
B Friction co-efficient between the plane and block is $0.5$
C Spring constant is $1000\text{N/m}$

Given, Inclination of plane, $\theta ={37}^{\circ }$
Mass of block, $m=2\text{kg}$


Block starts $4.8\text{m}$ away fromthe spring and compresses it by $0.2\text{m}$ and stops.

From Work Energy theorem
Work done by all forces = change in kinetic energy


Here, three forces operates - gravity, spring force and friction.
$W_g+W_s+W_f=0$
[Initial and final kinetic energy is zero]
$\Rightarrow mgd\sin \theta +(-\frac{1}{2}k{x}^2)-\mu mgd\cos \theta =0$ [spring is compressed, hence spring force will act upward and displacement is in downward direction, so work done by it is negative and friction always does negative work]
$\Rightarrow 2\times 10\times 5\times \sin {37}^{\circ }-\frac{1}{2}\times k\times (0.2{)}^2-\mu \times 2\times 10\times 5\times \cos {37}^{\circ }=0$
$\Rightarrow 60-0.02k-80\mu =0$ ... (1)


Block rebounds to $1\text{m}$ and stops.
$W_g+W_s+W_f=0$
[Initial and final kinetic energy is zero]
$\Rightarrow -mgd\sin \theta +\left(\frac{1}{2}k{x}^2\right)-\mu mgd\cos \theta =0$
[Spring force is acting upeard and displacement is also in upward direction. so workdone by spring is positive. Friction always do negative work and work done by gravitation is negative as block rebounds]
$\Rightarrow -2\times 10\times 1\times \sin {37}^{\circ }+\frac{1}{2}k(0.2{)}^2-\mu \times 2\times 10\times 1\times \cos {37}^{\circ }=0$
$\Rightarrow -12+0.02k-16\mu =0$ ... (2)

On adding equation (1) and (2)
$48-96\mu =0\Rightarrow \mu =0.5$ ... (3)
On putting value of equation (3) in (1)
$k=1000\text{N/m}$
Thus, option B and C are correct.