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
3m=B
2m=C
1m=D
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
=>v2-0=2(10)(1)
=>v2=20
Kinetic energy at B=(1/2)mv2
=(1/2)(20)(20) J .......[v2=20 as proven above]
=200J
Potential energy at C =mgh=(20 x 10 x 2)J
=400J
For Kinetic energy at C
by using equation of motion-v2-u2=2as
=>v2-0=2 x 10 x 2
=>v2=40
Kinetic energy at C =(1/2)mv2
=(1/2)(20)(40) J .....[v2 = 40 as proven above]
=400J
Potential energy at D =mgh=(20 x 10 x 1)J
=200J
For Kinetic energy at D
By using equation of motion- v2 - u2=2as
=>v2 - 0=2 x 10 x 3
=>v2=60
Kinetic energy at D =(1/2)mv2
=(1/2)(20)(60) J
=600 J
hope it helps ......:)