JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Laws of Motion

Two blocks

A

and

B

of masses

m_{A} = 1 kg

and

m_{B} = 3 kg

are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force

F

that can be applied on

B

horizontal, so that the block A does not slide over the block B is:
[Take

g = 10 m / s^{2}

]

(A)8 N
(B)16 N✓ Correct
(C)12 N
(D)40 N

Answer: (B)

Solution

Block

A

rides on

B

only as long as friction can supply its acceleration. The most friction

A

can feel is

f = μ m_{A} g = 0.2 \times 1 \times 10 = 2 N

, which caps the shared acceleration at

a = \frac{f}{m _{A}} = 2 m s^{- 2}

. Treating

A

and

B

as one body of mass

4 kg

, the table drags back with

μ (m_{A} + m_{B}) g = 0.2 \times 4 \times 10 = 8 N

, so Newton's second law reads

F - 8 = 4 \times 2

, giving

F = 16 N

as the largest force that keeps them moving together.

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Laws of Motion

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