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A block of mass 4 kg is placed on the fl...

A block of mass 4 kg is placed on the floor. The coefficient of static friction is 0.4. If a force of 12 N is applied on the block parallel to the floor, the force of friction between the block and floor `(g = 10 ms^(-2))`

A

zero

B

8N

C

12N

D

16 N

Text Solution

Verified by Experts

The correct Answer is:
d
Force of limiting friction, `F_(f)= m_(s) mg`
`F_(f) = 0.4 xx 4 xx 10 = 16 N`
Since applied force 12 N is less than limiting friction therefore, force of friction will be 12 N as friction will be 12 N as friction is a self adjusting force upon `F_(f)` Hence correct choice is (c)
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