Question

The work required to increases the velocity of a particle from 18 km/h to 72 km/h, if mass of particle is 2 kg, is

Answer 1

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v=72 km/h=20m/s ,u=18 km/h=5m/s
work = 1/2mv^2 - 1/2 mu^2
= 1/2*2*20^2 - 1/2*2*5^2
= 20^2 - 5^2
=400 - 25
=375 joules.
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Answer 2

Answer:

375J,Plz mark me as the brainliest,and thank me if it helped :)

Explanation:

Kinetic energy of an object =1/2*m*v^2

Work=Change in kinetic energy=(Final kin. energy-Initial kin. energy)

Remember,all values should be in SI units(That's where most of us make mistakes xD).So,

Initial velocity=18kmph=5m/s

Final velocity=72km/h=20m/s

Mass=2kg

Initial kinetic energy=1/2*2*25=25J

Final kinetic energy=1/2*2*400=400J

Work=Change in kinetic energy=(Final kin. energy-Initial kin. energy)=400-25=375J

Hope it helped,plz mark it as the brainliest ans :)