Question

Figure (8-E12) shows two blocks A and B, each of mass of 320 g connected by a light string passing over a smooth light pulley. The horizontal surface on which the block A can slide is smooth. Block A is attached to a spring of spring constant 40 N/m whose other end is fixed to a support 40 cm above the horizontal surface. Initially, the spring is vertical and unstretched when the system is released to move. Find the velocity of the block A at the instant it breaks off the surface below it. Take g = 10 m/s².

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Mass}\mathrm{of}\mathrm{each}\mathrm{block},m=320g=0.32\mathrm{kg} \\ \mathrm{Spring}\mathrm{constant},k=40\mathrm{N}/\mathrm{m} \\ h=40\mathrm{cm}=0.4m\mathrm{and}g=10\mathrm{m}/s^2 \end{gathered}$$



From the free-body diagram,
$kx\cos \mathrm{\theta }=mg$
As, when the block breaks of the surface below it (i.e. gets dettached from the surface) then R =0.

$$\begin{gathered} \Rightarrow \cos \mathrm{\theta }=\frac{mg}{kx} \\ \Rightarrow \frac{0.4}{0.4+x}=\frac{3.2}{40x} \\ \Rightarrow 16x=3.2x+1.28 \\ \Rightarrow x=0.1\mathrm{m} \\ \mathrm{So},s=\mathrm{AB}=\sqrt{\left(h+x^2\right)-h^2} \\ =\sqrt{{\left(0.5\right)}^2-{\left(0.4\right)}^2}=0.3\mathrm{m} \end{gathered}$$

Let the velocity of body B be $\nu$.
Change in K.E. = Work done (for the system)

$$\begin{gathered} \left(\frac{1}{2}m{u}^2+\frac{1}{2}m{\nu }^2\right)=-\frac{1}{2}k{x}^2+mgs \\ \Rightarrow \left(0.32\right)\times {\nu }^2 \\ =-\left(\frac{1}{2}\right)\times 40\times {\left(1.0\right)}^2+\left(0.32\right)\times 10\times \left(0.3\right) \\ \Rightarrow \nu =1.5\mathrm{m}/s \end{gathered}$$

From the figure,
$∆l=h\left(\sec \mathrm{\theta }-1\right)...\left(i\right)$

From the principle of conservation of energy,
$$\begin{gathered} mgs=2\left[\frac{1}{2}m{\nu }^2\right]+\frac{1}{2}K∆l^2 \\ mgh\tan \mathrm{\theta }=m{\nu }^2+\frac{1}{2}k{h}^2{\left(\sec \mathrm{\theta }-1\right)}^2...\left(ii\right) \end{gathered}$$

When the motion of the block breaks of the surface below it (i.e gets dettached from the surface on which it was initially placed) then
$$\begin{gathered} mg=kh\left(\sec \mathrm{\theta }-1\right)\cos \mathrm{\theta } \\ \Rightarrow 1-\cos \mathrm{\theta }=\frac{\mathit{m}\mathit{g}}{\mathit{k}\mathit{h}} \\ \Rightarrow \cos \mathrm{\theta }=1-\frac{\mathit{m}\mathit{g}}{\mathit{k}\mathit{h}} \\ \mathrm{or}\cos \mathrm{\theta }=\frac{kh-mg}{kh} \\ =\frac{40\times 0.4-0.32\times 10}{40\times 0.4} \\ =0.8 \end{gathered}$$

Putting the value of θ in equation (ii), we get:
$0.32\times 10\times 0.4\times 0.75$

$$\begin{gathered} =0.32{\nu }^2+\frac{1}{2}40\times {\left(0.4\right)}^2{\left(1.25-1\right)}^2 \\ \Rightarrow 0.96=0.32{\nu }^2+0.2 \\ \Rightarrow 0.32{\nu }^2=0.72 \\ \Rightarrow \nu =1.5\mathrm{m}/\mathrm{s} \end{gathered}$$