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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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Two blocks are resting on a floor as shown. There is no friction between floor and lower block (M=3 kg) but frictional coefficient between both blocks is 0.2. Both blocks move together with initial speed V towards the spring ,compresses it and due to force exerted by spring moves in reverse direction of initial motion. What is maximum value of V in (m/s), so that during motion there is no slipping between the blocks.


a) 20 m/s b) 10 m/s c) 30 m/s d)40 m/s
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1 Answer

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Maximum change of slipping can occur when the spring is compressed maximum.
Force provided by spring $=kx_m$
Acceleration of system of two mass $=\large\frac{kx_m}{M+m}$
For not slipping
Frictional force between two blocks must be greater than $ma$
$\mu mg >ma$
$\mu mg >\large\frac{kx_mm}{M+m}$-----(1)
Using work energy theorem for system
$x_m=\sqrt {\large\frac{(M+m)v^2}{k}}$-----(2)
Substituting for $x_m$ in (1) we get, limiting values
$\mu mg=\large\frac{kV \sqrt {\Large\frac{M+m}{k}}m}{M+m}$
Therefore $v=\mu g\sqrt {\large\frac{M+m}{k}}$
$\qquad=.2 \times 10 \sqrt {\large\frac{3+1}{400}}$
Pulling values for $\mu,g,M\; and\;m$
we get $v=20\;m/s$
Hence a is the correct answer.
answered Jul 19, 2013 by meena.p
edited Jun 1, 2014 by lmohan717

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