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.
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a max formula applied and f=mug applied
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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)
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