Question

Figure shows a spring 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 $5.8\text{m}$ away 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. The friction coefficient between the plane and the block and the spring constant of the spring is (Take $g=10{\text{m/s}}^2$)

• A. $\mu =0.54,k=1032\text{N/m}$
• B. $\mu =0.12,k=1032\text{N/m}$
• C. $\mu =0.12,k=962\text{N/m}$
• D. $\mu =0.28,k=962\text{N/m}$

Answer 1

The correct option is A $\mu =0.54,k=1032\text{N/m}$

FBD of block for downward and upward motion along the incline:


Resolving all forces along the plane, we get
Net force $=mg\sin {37}^{\circ }-\mu N$
$=mg\sin {37}^{\circ }-\mu mg\cos {37}^{\circ }$
Distance covered along the plane $=5.8\text{m}+0.2\text{m}=6.0\text{m}$
So, work done by all forces on the block $=F\times d$
$=(mg\sin {37}^{\circ }-\mu mg\cos {37}^{\circ })\times 6.0$
... (1)

This work done is stored as potential energy in the spring compression $=\frac{1}{2}k{x}^2=\frac{1}{2}k(0.2{)}^2$ ... (2)

When block rebounds, this P.E is utilized in the motion of the block. In return (rebound) journey, the friction force remains same, but acts in opposite direction of motion.
$\therefore$ New net force along the plane
$=mg\sin {37}^{\circ }+\mu mg\cos {37}^{\circ }$
Now it moves for a distance of $1\text{m}$
$\therefore$ Work done by forces $=(mg\sin {37}^{\circ }+\mu mg\cos {37}^{\circ })\times 1.0$ ... (3)

(a) Equating (1) and (3):
$(mg\sin {37}^{\circ }+\mu mg\cos {37}^{\circ })=6(mg\sin {37}^{\circ }-\mu mg\cos {37}^{\circ })$
$\Rightarrow 7\mu mg\cos {37}^{\circ }=5mg\sin {37}^{\circ }$
$\Rightarrow \mu =\frac{5mg\sin {37}^{\circ }}{7mg\cos {37}^{\circ }}$
$\Rightarrow \mu =\frac{5}{7}\tan {37}^{\circ }=0.54$
$\therefore \mu =0.54$

(b) Equating (2) and (3), we get
$\frac{1}{2}k{x}^2=mg\sin {37}^{\circ }+\mu mg\cos {37}^{\circ }$
$\Rightarrow k(0.2{)}^2=2mg(\sin {37}^{\circ }+\mu \cos {37}^{\circ })$
$\Rightarrow k=\frac{2\times 2\times 10(0.6+0.54\times 0.8)}{0.04}$
$\Rightarrow k=\frac{40\times 1.032}{0.04}=1032\text{N/m}$
$\therefore$ Stiffness $k=1032\text{N/m}$