Question

A body of mass 20kg is dropped from a height of 100m. Find its kinetic energy and potential energy after 1sec,2sec,3sec

Answer 1

(1) When an object fall then initially the object is at rest, therefore using equation of motion we have,
 v=u+gtor,v=0+gtv=gt

For 1st second, the final velocity of the object will be,
 v=9.8 m/s2×1 sv=9.8 m/s
Therefore the kinetic energy for the first second will be,
 KE1=12mv2or,KE1=12mg2t2
On substituting the values we get,
KE1=12×20 kg×(9.8 m/s2g)2(1 s)2KE1=960.4 kg.m2/s2KE1=960.4 Joule
 
And the distance fallen by the object will be given by using equation of motion as,
 h1=ut+12gt2or,h1=0+12gt2h1=12gt2
Thus, the potential energy at 1st second will be,
 PE1=mgh1
On substituting the values we get,
 PE1=20 kg×9.8 m/s2×129.8 m/s2×(1s)2PE1=960.4 kg.m2/s2PE1=960.4 Joule

Similarly, you can calculate for 2nd second and for the 3rd second