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Work and Energy — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsWork and Energy5 Marks
Question
Figure shows two blocks $A$ and $B,$ each having a mass of $320g$ 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 $40N/m$ 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 $g = 10m/s^2.$
✓

Answer

$m = 320g = 0.32\ kg, k = 40N/m h = 40\ cm = 0.4m, g = 10 m/s^2$ From the free body diagram,

$\text{kx}\cos\theta=\text{mg}$
$($when the block breaks off $R = 0)$
$\Rightarrow\cos\theta=\frac{\text{mg}}{\text{kx}}$
So, $\frac{0.4}{0.4+\text{x}}=\frac{3.2}{40+\text{x}}$
$\Rightarrow16\text{x}=3.2\text{x}+1.28$
$\Rightarrow\text{x}=0.1\text{m}$
So, $\text{s}=\text{AB}$
$\Rightarrow\sqrt{(\text{h}+\text{x})^2-\text{h}^2}$
$\Rightarrow\sqrt{(0.5)^2-(0.4)^2}=0.3\text{m}$
Let the velocity of the body at $B$ be $v,$ Charge in $K.E. =$ work done $($for the system$)\Big(\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{mv}^2\Big)=\frac{-1}{2}\text{kx}^2+\text{mgs}$
$\Rightarrow(0.32)\times\text{v}^2=-\Big(\frac{1}{2}\Big)\times40\times(0.1)^2+0.32\times10\times(0.3)$
$\Rightarrow\text{v}=1.5\text{m}/\text{sec}$

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