Question

A body of mass 20kg is dropped from a height of 100m.Find its K.E and P.E after

(i)1st second

(ii)2nd second

(iii)3rd second

Answer 1

(1) equation of v = u gt
or, v = 0 gt v = gt  final velocity of the object will be, v = 9. 8 m/s2 x 1s v= 9. 8 m/s
 Therefore the kinetic energy for the first second will be,
 KE1 = 2 rirV2 Or, , KE1 = 2x 20 kg x (9.8 m/s2g)2(1 s)2 KE1 = 960.4 kg. m2 / KE1 = 960.4 Joule And the distance fallen by the object will be given by using equation of motion as, h1 = ut + 4gt2 or, s 1„a2 And the distance fallen by the object will be given by using equation
 h1= ut bt2 or, h1 = 0 + *gt2
hi = *02
Thus, the potential energy at 1st second will be
, PEI = mghl On substituting the values we get,
 PEJ. = 20 kg x 9.8 m/s2 x 49.8 m/s2 x (1s)2
PEI = 960.4 kg. m2 / s2 PEI = 960.4 Joule Similarly, you can calculate for 2nd second and for the 3rd second. 


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