Physics, asked by bhaifake03, 1 month ago

Derive an expression K.E = ½ mv^2 for a freely falling body of mass m

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Answers

Answered by rai64603
0

Answer:

Kinetic Energy:

The energy possessed by an object due to its motion is called as kinetic energy.

Derivation:

Let us consider an object of mass "m" which is at rest on smooth horizontal plane.

Let a Force ,F acts on the object and let the object from rest moves from point A to point B and covers a displacement S.

The Work done by Force on the object is :

Workdone = Force x displacement.

W=F×S ____(1).

From third equation of motion;

V

2

−U

2

=2aS

S=

2a

V

2

−U

2

______(2)

By Newtons second law of motion:

F=ma

From equation 1 and 2

W=

2a

m∗a∗(V

2

−U

2

)

As we have assumed object is at rest, U=0

W=

2

mV

2

The Work Done appears as kinetic energy of the body.

Therefore,

K.E=(

2

1

)mV

2

Answered by ItzUnlcukygirl
4

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Kinetic Energy:

The energy possessed by an object due to its motion is called as kinetic Energy.

Derivation:

Let us consider an object of mass " m " which is at rest on smooth horizontal plane.

Let a Force , F acts on the object and let the object from rest moves from point A to point B and covers a displacement S.

The WORK Done by Force on the object is :

Workdone = Force x displacement.

W= FxS ____(1).

From third equation of motion;

V^2 -U^2 =2aS

S= V^2 -U^2/2a ______(2)

By Newtons second law of motion:

F= ma

From equation 1 and 2

W= m*a* (V^2- U^2)/2a

As we have assumed object is at rest, u=0

W=m*V^2/2

The WORK Done appears as kinetic energy of the body.

Therefore,

K.E = (1/2 )mv^2

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