JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Laws of Motion

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The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is _____ N. (take

g = 10 ms^{- 2}

)

Answer: 15

Solution

The horizontal force acts on the lower

2

kg block, and the upper

1

kg block is carried along only by friction. For the upper block,

f = (1 kg) a \leq μ (1 kg) g = 0.5 \times 1 \times 10 = 5 N

so the largest acceleration the pair can share is

a = 5 m s^{- 2}

. The table is smooth, so the applied force has only to accelerate the full

3

kg:

F_{m a x} = (1 + 2) (5) = 15 N

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Integer answer

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