__Question 6.20__:^{3/2}where a = 5 m

^{1/2}s

^{-1}. What is the work done by the net force during its displacement from

*x*= 0 to

*x*= 2 m?

__Solution__:Mass of the body,

*m*= 0.5 kg

^{3/2}and a = 5 m

^{1/2}s

^{-1}

Initial velocity,

*u*(at

*x*= 0) = 0

Final velocity

*v*(at

*x*= 2 m) = 10√2 m/s

Work done,

*W*= Change in kinetic energy

= (1/2) m (v

^{2}- u

^{2})

= (1/2) X 0.5 [ (10√2)

^{2}- 0

^{2}]

= (1/2) X 0.5 X 10 X 10 X 2

= 50 J

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