Question

3. - A block of weight 20 N is pressed against a vertical wall as shown in figure by applying a horizontal force of the block (g = 10 ms?). If , = 0.4 then The least horizontal force required to keep the block in equilibrium is a) 40 N b) 20 N c) 10 N d) 50 N F

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Answer 1

the answer of this question is 40n

Answer 2

d)50N

Explanation,

Given,

• A block of weight 20N

• Coefficient of friction = 0.4

To find,

The least horizontal force required to keep the block in equilibrium.

Solution,

• The weight of the block will be acting in the downward direction.

• The friction between the wall and block will act in the upward direction.

• The direction of frictional force is always in such a way that it opposes the motion.

Frictional force = friction coefficient* normal reaction.

Here, the normal reaction will be equal to the horizontal force being applied.

Let the horizontal force be $x$.

Frictional force = $0.4x$

Frictional force balances the weight, then according to the question,

          Frictional force =  Weight of the body

                               $20N = 0.4x\\x = 50N$

Therefore, $50N$ horizontal force is required to keep the block in equilibrium.