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Question -

┬аA body of mass 2 kg initially at rest movesunder the action of an applied horizontal force of 7 N on a table withcoefficient of kinetic friction = 0.1. Compute the
(a) Work done by the appliedforce in 10 s
(b) Work done by friction in 10 s
(c) Work done by the net force onthe body in 10 s
(d) Change in kinetic energy ofthe body in 10 s and interpret your results.



Answer -

(a) We know that Uk┬а=frictional force/normal reaction
frictional force = Uk┬аxnormal reaction
= 0.1 x 2 kg wt = 0.1 x 2 x 9.8 N = 1.96 N
net effective force = (7 тАУ 1.96) N = 5.04 N
acceleration = 5.04/2 ms-2┬а=2.52 ms-2
distance, s=1/2x 2.52 x 10 x 10 = 126 m
work done by applied force = 7 x 126 J = 882 J
(b) Work done by friction = 1.96 x 126 = -246.96 J
(c) Work done by net force = 5.04 x 126 = 635.04 J
(d) Change in the kinetic energy of the body
= work done by the net force in 10 seconds = 635.04 J (This is in accordancewith work-energy theorem).

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