Question

The work required to increase the velocity of a particle from 18 km per hour to 72 km per hour if mass of particle is 2 kg is

Answer 1

Kinetic energy= 1/2×m×v²

K.E(Initial) = 1/2×2×(18×5/18)²  {convenrting km/h to m/s}

K.E(Initial) = 25 J

Kinetic energy= 1/2× 2 ×(72×5/18)²

Kinetic energy= 400 J

Work done= Final K.E - Initial K.E

                  = 400 - 25

                  = 375 J

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 :)