Question

Figure (8-E7) shows a spring fixed at the bottom end of an incline of inclination 37°. 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 m up the incline. Find (a) the friction coefficient between the plane and the block and (b) the spring constant of the spring. Take g = 10 m/s².
Figure

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=2\mathrm{kg} \\ \mathrm{Initial}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{from}\mathrm{the}\mathrm{spring},S_1=4.8\mathrm{m}, \\ \mathrm{Comression}\mathrm{in}\mathrm{the}\mathrm{spring},x=20\mathrm{cm}=0.2\mathrm{m} \\ \mathrm{Final}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{block}\mathrm{from}\mathrm{the}\mathrm{spring},{\mathrm{S}}_2=1\mathrm{m} \\ \mathrm{As}\mathrm{\theta }=37°, \\ \sin 37°=0.60=\frac{3}{5} \\ \cos 37°=0.80=\frac{4}{5} \end{gathered}$$

Applying the work-energy principle for downward motion of the block,

$$\begin{gathered} 0-0=mg\sin 37°\left(x+4.8\right)-\mu \mathrm{R}\times 5-\frac{1}{2}k{x}^2 \\ \Rightarrow 20\times \left(0.06\right)\times 5-\mu \times 20\times \left(0.80\right)\times 5-\frac{1}{2}k{\left(0.2\right)}^2=0 \\ \Rightarrow 60-80\mu -0.02k=0 \\ \Rightarrow 80\mu +0.02k=60...\left(i\right) \end{gathered}$$

Similarly for the upward motion of the body the equation is
$$\begin{gathered} 0-0=\left(-mg\sin 37°\right)-\mu \mathrm{R}\times 1+\frac{1}{2}k{\left(-2\right)}^2 \\ \Rightarrow 20\times \left(0.06\right)\times 1-\mu \times 20\times \left(0.80\right)\times 1-\frac{1}{2}k{\left(0.2\right)}^2 \\ \Rightarrow 12-16\mu +0.02k=0 \end{gathered}$$

Adding equations (i) and (ii), we get:

$$\begin{gathered} 96\mu =48 \\ \Rightarrow \mu =0.5 \end{gathered}$$

Now putting the value of $\mu$ in equation (i), we get:
k = 1000 N/m