Laws of MotionHard
Question
A small block of mass 20 kg rests on a bigger block of mass 30 kg which lies on a smooth horizontal plane. Initially the whole system is at rest. The coefficient of friction between the blocks is 0.5 and a horizontal force of 50 N is applied on the lower block then work done by frictional force on upper block in t = 2 sec is -(surface is smooth )
Options
A.40 J
B.-40 J
C.80 J
D.-80 J
Solution
a = 1 m/s2The two blocks moves together with acceleration
a =
Friction force on upper block
= f = ma = 20 × 1 = 20 N
fmax = μmg = 0.5 × 20 × 10 = 100 N
Thus no slipping occurs
Distance travelled = S = ut +
at2 = 0 + 
× 1 × (2)2 = 2mWork =
= (F) (S) cos 0o= (20) (2) (1) = 40 J
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