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Question

Two blocks (m=0.5 kg and M=4.5 kg) are arranged on a horizontal frictionless table as shown in the figure. The coefficient of static friction between the two blocks is 3/7. Then the maximum horizontal force that can be applied on the larger block so that blocks move together is ______N.

(Round off to the nearest integer)
(Take g as 9.8 ms2)



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Solution

Maximum acceleration for both blocks to move together,
amax=μg
amax=37×9.8

amax=4.2 m/s2

Now, considering both blocks are moving together with the acceleration of 4.2 m/s2,

Fmax=(M+m)amax

Fmax=(4.5+0.5)×4.2=21 N

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