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System shown in figure is released from rest. Pulley and spring is massless and friction is absent everywhere. The speed of 5kg block when 2 kg block leaves the contact with the ground is $(k=40 N/m;g=10 m/s^2)$

\[(a)\;\sqrt 2\;m/s \quad (b)\;2 \sqrt 2\;m/s \quad (c)\; 2\; m/s \quad (d)\;4 \sqrt 2\;m/s\]

1 Answer

Let x be the extension in spring when 2 kg mass leaves contact with the ground.
$kx=2 g$
$\quad=\large\frac{2 \times 10}{40}=\frac{1}{2}$$m$
By law of conservation of energy for 5 kg mass
$mgx=\large\frac{1}{2}$$ k^2+\large\frac{1}{2}$$mv^2$
$v= \sqrt {2gx-\large\frac{kx^2}{m}}$
$v=\sqrt {2 \times10 \times \large\frac{1}{2} -\large\frac{40}{5} \times \large\frac{1}{4}}$
$v=2 \sqrt 2 m/s$
Hence b is the correct answer.
answered Aug 5, 2013 by meena.p
edited Jun 6, 2014 by lmohan717

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