JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Laws of Motion

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The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall be ______ N.
[

g = 10

^{- 2}

]

Answer: 25

Solution

For the block to remain stationary against the vertical wall, the friction force must balance the weight:

f = m g

The maximum static friction is:

f_{ma x} = μ N

where

N

is the normal force (equal to the applied horizontal force

F

).

For equilibrium:

μ F = m g

F = \frac{m g}{μ} = \frac{0.5 \times 10}{0.2} = \frac{5}{0.2} = 25 N

Answer: 25

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Integer answer

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