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)mv^{2}

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- v^{2}- u^{2}=2as

=>v^{2}-0=2(10)(1)

=>v^{2}=20

Kinetic energy at B=(1/2)mv^{2}

=(1/2)(20)(20) J .......[v^{2}=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-v^{2}-u^{2}=2as

=>v^{2}-0=2 x 10 x 2

=>v2=40

Kinetic energy at C =(1/2)mv^{2}

=(1/2)(20)(40) J .....[v^{2} = 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- v^{2} - u^{2}=2as

=>v^{2} - 0=2 x 10 x 3

=>v^{2}=60

Kinetic energy at D =(1/2)mv^{2}

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

=600 J

hope it helps ......:)

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