Question

50 points,
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plz solve for class 11th
friction.
Two blocks of mass 4 kg and 6 kg are placed one over the other. If a force of 10 N is applied on the block of mass 4 kg, then the accelerations of the two blocks at the instant shown is
(g = 10 m/s²)
(A) (4 kg) = a(6 kg) = 0
(B) a(4 kg) = 0.5 m/s² a(6 kg) = 0
(C) a(4 kg) = 2 m/s² a(6 kg) = 0.5 m/s² (D) a[4 kg) = 1 m/s² a(6 kg) = 1 m/s²
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Answer 1

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CORRECT ANSWER IS OPTION D

➡️Draw FBD of two blocks

➡️ assuming that blocks moves with common acceleration

ac = 10/4+6 = 10/10= 1m/s²

➡️draw FBD of block b (on which external force is not applied )

➡️ Required frictional force

f = Mb × ac = 6× 1 = 6N

since, limiting friction (maximum available static friction)=fL=μN1 0.2×4×10 =8N

NOTE:- If ac (common acceleration) is ≤ a max then,

aA=aB=aCommon

(f is less then FL)Required frictional force is less then limiting friction hence, both blocks will move together with common acceleration.

Answer 2

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Your answer is option D!~ for full solution see the above attachment ❤

hope it helps you my dear friend!~♡