Question

If acceleration of A is 2m/s2which is smaller than acceleration of B then the value of frictional force applied by B on A is:-

a) 50N
b) 20N
c) 10N
d) None of these


Kindly answer sir/mam.

Answer 1

a max formula applied and f=mug applied

Answer 2

Answer: 10N

Explanation: As acceleration of A is smaller than B , so there must be a frictional coefficient working between the two.

So now , acceleration of A = (Mu ie. Frictional coefficient) × g

acceleration of A = 2 (given)

Put g =10

So u will find Mu = 0.2

Now the frictional force applied by B on A = Mu × (Normal force which is nothing but mA ×g)

Therefore F= 0.2 ×(5×10)

So F=10N (ans)