Question

The 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 $4.8\text{m}$ 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. Find the friction coefficient between the plane and the block and the spring constant of the spring.
Take $g=10{\text{m/s}}^2$.

• A. $0.25,100\text{N/m}$
• B. $0.25,1000\text{N/m}$
• C. $0.5,100\text{N/m}$
• D. $0.5,1000\text{N/m}$

Answer 1

The correct option is D $0.5,1000\text{N/m}$

Given:
$m=2\text{kg};s_1=4.8\text{m};s_2=1\text{m}$
$x=20\text{cm}=0.2\text{m};g=10{\text{m/s}}^2$

Applying work- energy principle for downward motion of the body between initial and final point where velocity of block become zero.

$K{E}_f-K{E}_i=W_{\text{all forces}}$

$K{E}_f-K{E}_i=W_{\text{gravity}}+W_{\text{friction}}+W_{\text{spring}}$

$0-0=[mg\sin {37}^{\circ }\times (s_1+x\left)\right]+[-\mu mg\cos {37}^{\circ }(s_1+x\left)\right]+[-\frac{1}{2}K{x}^2]$

$0=[20\times 0.6\times 5]+[-\mu \times 20\times 0.8\times 5]+[-\frac{1}{2}K\times {0.2}^2]$

$60-80\mu -0.02K=0$

$80\mu +0.02K=60.....\left(1\right)$

Similarly, for the upward motion of the body the equation is

$0-0=[-mg\sin {37}^{\circ }\times (s_2\left)\right]+[-\mu mg\cos {37}^{\circ }(s_2\left)\right]+[-\frac{1}{2}K{x}^2]$

$0=[-20\times 0.6\times 1]+[-\mu \times 20\times 0.8\times 1]+[\frac{1}{2}K\times {0.2}^2]$

$16\mu -0.02K=-12.....\left(2\right)$

Adding equation $\left(1\right)$ & equation $\left(2\right)$, we get

$96\mu =48$

$\therefore \mu =0.5$

Now putting the value of $\mu$ in equation $\left(1\right)$, we get

$K=1000\text{N/m}$

Hence, option $\left(D\right)$ is correct.