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Consider the system shown below A ho...

Consider the system shown below

A horizontal force F is applied to a block X of mass 8 kg such that the block Y of mass 2 kg adjacent to it does not slip downwards under gravity. There is no friction between the horizontal plane and the base of the block X. The coefficient of friction between the surfaces of blocks X and Y is 0.5. Take acceleration due to gravity to be `10 ms^(-2)`. The minimum value of F is

A

200N

B

160N

C

40N

D

240N

Text Solution

Verified by Experts

The correct Answer is:
A
FBD of y
Acceleration of system
`a = (F)/(8 + 2) = (F)/(10)`
At equilibrium
N = ma
`f = mg implies mu N = mg`
`implies 0.5 ma = mg implies 0.5 xx (F)/(10) = 10`
`implies F = 200 N`
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