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Figure shows a small block of mass m kept at the left end of a larger block of mass M and length l.The system can slide on a horizontal road. The system is started towards right with an intial velocity v. The friction coefficient between the road and the bigger block is µ and that between the blocks is µ/2.Find the time elapsed before the smaller block separates from the bigger block.

Figure shows a small block of mass m kept at the left end of a larger block of mass M and length l.The system can slide on a horizontal road. The system is started towards right with an intial velocity v. The friction coefficient between the road and the bigger block is µ and that between the blocks is µ/2.Find the time elapsed before the smaller block separates from the bigger block.

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Grade:11

2 Answers

Anonymous
11 Points
6 years ago
Start with drawing the free body diagrams of both the blocks. Let the smaller block be 2 and bigger be 1 and same for their acceleration. Now, the forces acting on 1 are the frictional force applied by the ground which is f_{1}= \mu *(M+m)*g and the reaction force of the friction between 1 and 2 which is f_{2}= \mu *m*g in the forward direction(since it is a reaction force). Similarly the force on 2 will be just f_{2}= \mu *m*g in the backward direction. You will get the retardation rate of both  the blocks let it be a1 and a2 respectively. now calculate the relative acc. of 2 w.r.t 1  a_{r}= a_{2}-a_{1} which will be a_{r}= \frac{(\mu *(M+m)*g)}{2*m}. Now use Newton’s Second equation of motion and get the time. Also there might be quetion on whether the block will fall or not because of the velocity. You might have to find minimum velocity required for the block to fall which can be calculated based on the time required. As the key to fall is the relative acceleration, as soon as 1 stops, friction will stop the other one too, you would need the velocity of 2 at that time based on which you can do the calculation using the remaining distance and retardation which you know.
Hope it helped!
Vikas TU
14149 Points
6 years ago
Dear Student,
Balancing the friction forces and eqns. we get,
f1=µmg/2
f2=µMcm/g
a1=-µg/2
a2=-µMcmg/M
relative a= a1-a2
=-µg/2 + µMcmg/M
relative distance=h
vrel=0
h=relative a x t2/2
=>t=√(2ML/(2µmg+µMg))
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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