An object of mass of 20 kg is dropped from a height of 4 m.  Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case.

height at which object is located:  4 m, 3 m, 2m, 1m, just above the ground

Potential energy

Kinetic energy

(for simplifying the calculations, take the value of g as 10 m s-2)

Lets call height at 4m=A




Potenial energy a A = mgh=(20 x 10 x 4 )J=800 J 

Kinetic energy at A =(1/2)mv2

as we know body is stationary at A so v=0 so K.E.=0

Potential energy at B =mgh=(20 x 10 x 3)J =600J

For Kinetic energy at B

By equation off motion- v2- u2=2as



Kinetic energy at B=(1/2)mv2

=(1/2)(20)(20)   J   .......[v2=20 as proven above]


Potential energy at C =mgh=(20 x 10 x 2)J


For Kinetic energy at C

by using equation of motion-v2-u2=2as

=>v2-0=2 x 10 x 2


Kinetic energy at C =(1/2)mv2

=(1/2)(20)(40) J     .....[v2 = 40   as proven above]


Potential energy at D =mgh=(20 x 10 x 1)J


For Kinetic energy at D

By using equation of motion- v2 - u2=2as

=>v2 - 0=2 x 10 x 3


Kinetic energy at D =(1/2)mv2

=(1/2)(20)(60) J

=600 J

hope it helps ......:)

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