Question

Figure (8-E7) shows a spring fixed at the bottom end of Ae:.. an incline of inclination. ${37}^{\circ }$ A small block of mass 2 kg starts slipping down the incline from a point 4.8 m away from the spring. The block compresses the spring by 20 cm, stops momentarily and then rebounds through a distance of 1 in up the incline. Find (a) the friction : coefficient between the plane and the block and (b) the v spring constant of the spring. Take$g=10m/s^2$

Answer 1

$m=2kg,S_1=4.8m,x=20cm=0.2m$

$S_2=1m,$

$sin{37}^{\circ }=0.60=\frac{3}{5}$

$\theta ={37}^{\circ }$

$cos{37}^{\circ }=0.80=\frac{4}{5}=10m/se{c}^2$

Applying, work energy principle for downward motion of the body,

$0-0=mgsin{37}^{\circ }(x+4.8)-\mu R\times 5-\frac{1}{2}k{x}^2$

$20\times \left(0.06\right)\times 5-\mu \times 20$

$\Rightarrow \times \left(0.80\right)\times 5-\frac{1}{2}k(0.2{)}^2=0$

$\Rightarrow 60-80\mu -0.02k=0$

$\Rightarrow 80\mu +0.02k=60$ ...(i)

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

$0-0=(-mgsin{37}^{\circ })-\mu R\times 1+\frac{1}{2k}(-2{)}^2$

$\Rightarrow 20\times \left(0.06\right)\times 1-\mu \times 20\times \left(0.80\right)\times 1-\frac{1}{2}k(0.2{)}^2$

$\Rightarrow 12-16\mu +0.02k=0$

Adding equation (i) and equation (ii), we get ,

$96\mu =48$

$\Rightarrow \mu =0.5$

Now putting the value of $\mu$ in equation (i),

$k=1000$ N/m