Question

A body of mass 1 kg initially at rest is moved by a horizontal for force of 0.5 N on a smooth table.Calculate the work done by the force in 10 seconds and show that it is equal to the change in K.E of the body.​

Answer 1

0.5 = a

s = ut + 1/2at^2

s =( 1/2 ) (0.5) 100

s = 25

work done = 0.5 × 25 = 12.5

v^2 = u^2 + 2as

v^2 = 2 × 0.5 × 25 = 25

change in kinetic energy = 1/2 m v ^2

=1/2 × 1 × 25

= 12.5

Hopes it helps !!!

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