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Solve the following :

Question:

Figure (8-E12) shows two blocks $A$ and $B$, each having a mass of $320 \mathrm{~g}$ connected by a light string passing over a smooth light pulley. The horizontal surface on which the block $A$ can slide is smooth. The block $A$ is attached to a spring of spring constant $40 \mathrm{~N} / \mathrm{m}$ whose other end is fixed to a support 40 $\mathrm{cm}$ above the horizontal surface. Initially, the spring is vertical and upstretched 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 $g=10 \mathrm{~m} / \mathrm{s}^{2}$.

Solution:

from figure

$\mathrm{kx} \cos \theta=\mathrm{mg}$

$\mathrm{kx} \cos \theta=m g$

$40 \times X \times(0.4 /[0.4+\times])=0.32 \times 10$

$\mathrm{x}=0.1 \mathrm{~m}$

$\mathrm{S}=0.3 \mathrm{~m}$

Change in K.E. = Work done

$\frac{1}{2} m v^{2}=-\frac{1}{2} k x^{2}+m g s$

$\frac{1}{2} \times 0.32 \times v^{2}=-\frac{1}{2} \times 40 \times 0.1^{2}+0.32 \times 10 \times 0.3$

$v=1.5 \mathrm{~m} / \mathrm{s}$

 

 

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