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
Create a free account to view solution
View Solution FreeMore Laws of Motion Questions
Two beads A and B move along a semicircular wire frame as shown in figure. The beads are connected by an inelastic strin...When a constant force is applied to a body, it moves with uniform :...Two blocks are in contact on a frictionless table. One has mass m and the other 2m. A force F is applied on 2m as shown ...Two masses m and M are attached with strings as shown. For the system to be in equilibrium we have...What is the maximum value of the force F such that the block shown in the arrangement, does not move :-...