Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Work, Power and Energy

Two masses $m_1$ and $m_2$ connected by light spring rests on horizontal plane. The coefficient of friction between the masses and surface is $\mu$. What minimum force has to be applied in horizontal direction to the mass $m_1$ in order to shift the mass $m_2$ $ (g= 10 m/s^2)$

$ a)\; F= \frac{m_1 \mu g}{2} \\ b)\; F=\bigg(m_1+\frac{m_2}{2}\bigg)\mu g \\ c)\; F= \frac{(m_1+m_2)}{2} \mu g \\ d) F=\bigg(\large\frac{m_1+m_2}{m_2}\bigg)\mu g $

1 Answer

Block $m_2$ will shift if $ kx=\mu m_2 g$-----(1)
For the mass $m_2$ equating the energies with the work done by force F
Work done by F=P.E of spring+work done against friction
$Fx=\large\frac{1}{2}$$kx^2+\mu m_1 x g$
$kx=2F-2 \mu m_1 g$-----(2)
Equating (1) and (2)
$F= \bigg(m_1+\large\frac{m_2}{2}\bigg)$$\mu g$
Hence b is the correct answer.


answered Aug 7, 2013 by meena.p
edited Feb 17, 2014 by meena.p

Related questions