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✓✓Answer this Fast!!... PLEASE ★ consider two observers one whose frame is attached to the ground and another whose frame is attached to the train moving with uniform velocity 'u'with respect to the ground. each observed that a particle initially at rest to the train is accelerated by a constant force applied to it for time 't' in the forward direction with an acceleration a then the gain in kinetic energy of the particle is measured by an observer in 1)the Train is 1/2ma²t² 2)The ground is 1/2ma²t² + maut 3)any frame is equal to work done by the force in that frame. 4)All of the above
Asked by jhajuhi19 | 23 Apr, 2019, 01:18: AM
(1) gain in kinetic energy seen by the observer at train is (1/2)ma2t2 .

Particle is initially at rest with respect to train and acclerated. Hence after time t its speed is  (a×t )  .
Hence kinetic energy (1/2) m (a×t)2 .  This is true

(2) gain in kinetic energy seen by the observer at train is (1/2)ma2t2 + maut

If particle is initially at rest with respect to train, then particle initial velocity with respect to ground is u.
After time t, due to acceleration, its speed will be (u+at). Hence kinetic energy = (1/2)m(u+at)2 = (1/2)mu2 + (1/2)ma2t2 + maut

Hence kinetic energy given in statement (2) is not matching, statement (2) is not correct

(3) workdone by the force on a particle will be seen as kinetic energy of the particle.
Since we have seen above, kinetic energy is not same in different frame, workdone also will not be same
Answered by Thiyagarajan K | 23 Apr, 2019, 10:30: AM

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