A rigid body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Calculate the
Text Solution
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875 Joule –250 joule
625 joule. 625 joule,
Change in kinetic energy of the body is equal to the work done by the net force in 10 second. This is in accordance with work-energy theorem.
Sol.
F – f = ma ⇒ a = 
= 
= 2.5 m/s 2
S = 
at 2 = 
× 2.5 × 100 = 125 m
W F = F × s = 7 × 125 = 875 J.
W f = –0.1 × 2 × 10 × 125 J = – 250 J
W Total = (875 – 250) = 625 J
K = W total = 625 J.
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