Question

a block of mass 2 kg is given v=12m/s² real life original surface coefficient of friction is 0.2 the magnitude and direction of frictional force acting on it after how much time and at what distance it will stop​

Answer 1

Answer:

Step 1: Draw free body diagram [Ref. Fig.]

Step 2: Finding friction

From figure, there is no acceleration in vertical direction. So, Applying Newton's Law in vertical direction, we get

N=mg

=50N

Maximum value of Friction f

max

=μN=0.2×50=10 N

Since, f

max

≤F

So, sliding will happen, and kinetic friction will act, i.e.

f

kinetic

=μN=10N

Step 3: Newton’s Law for finding acceleration

Applying Newton's second law on block, in horizontal direction

(Taking rightward positive)

∑F=ma

⇒F−f

kinetic

=ma

⇒F−μmg=ma

⇒40−0.2×5×10=5a

⇒a=6m/s

2

Hence acceleration of the block will be 6m/s

2

Hence, Option A is correct.