Question

Figure shows two blocks A and B, each having a mass of 320 gram 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 is attached

Answer 1

Mass of the Block $=320gm=0.32kg$

$g=10m{s}^{2}n=40cm=0.4mk=40N/m$

Kinetic energy =ot

Initial system=0

Let spring make an angle $\theta$ and force due to spring $T$ if velocity is $V$

Then Kinetic energy on Block $=2\times \frac{1}{2}m{c}^2=m{v}^2$

Now change in KE $=m{v}^2-0=m{v}^2$

Elongation in the spring

$T=kx$ where $x=ncosec\theta -n$

$T=K(ncosec\theta -n)$...........(1)

Now equation vertical Forces

$T\sin \theta =mg\Rightarrow T=mg/\sin \theta$

$\Rightarrow K(ncosec\theta -n)=mg/\sin \theta$ (from (1))

$\Rightarrow \sin \theta cosec\theta -\sin \theta =mg/kh$

$\Rightarrow 1-\sin \theta =mg/kh=(0.32\times 10)/(0.4\times 40)$

$\Rightarrow \sin \theta =1-3.2/16=1-0.2=0.8$

Now wlongation $x=4cosec\theta -4=\left(n/\sin \theta \right)-n=0.1$

$Displacements=n\cot \theta =0.4\times 3/4=0.3m$

$\therefore$ change in KE=work done by all Forces

$m{v}^2=mgs-\frac{1}{2}k{x}^2$ ($\because$ elongation in opp.dir)

$\Rightarrow v^2=2.375\Rightarrow v=\sqrt{2.375}=1.54m{s}^1$