Hi,

my name is Aprameya i had a doubt

we know K.E.(Kinetic Energy) = (1/2)mv^2

the v over here is velocity but at what time. Velocity will be different at differnt points of time i mean at 1second it might be something and and at another second it might be something else. so while calculating which velocity do we use

### Asked by aprameyachaubeycr7 | 23rd Mar, 2020, 02:01: AM

### Consider an example,
An object of mass 5 kg is at rest starts and then start moving with a velocity of 2 m/s and reaches point B from A.
A(u=0).................................B (v = 2m/s)
So here in this exampe if you are asked to find the kinetic energy at point A then,
K.E.A = 1/2 ×mu^{2 }
Here,
u - initial velocity
u =0 at point A
Thus
K.E at point A will be zero.
Now if you want to find kinetic energy at point B,
K.E. = 1/2 mv^{2}
v = 2m/s at point B
Thus,
K.E_{B} = 1/2 × 5 × 2^{2}
Thus,
K.E._{B} = 10 Joule
Now conisder another example if an object starts moving from point B and reaches point C and at point C the velocity is 4 m/s
Then the K.E. = 1/2 (5)(4)^{2} = 40 J
Also the change in the kinetic energy can be calculated using formula,
K.E. = 1/2 m (v^{2} - u^{2})
So in case of motion from B to C the change in kinetic energy is,
K.E. = K.E._{f} - K.E._{i}
K.E. = 1/2 (5) (4^{2} - 2^{2} ) = 30 J.
Hope this help you to understand.

^{2 }

^{2}

_{B}= 1/2 × 5 × 2

^{2}

_{B}= 10 Joule

^{2}= 40 J

^{2}- u

^{2})

_{f}- K.E.

_{i}

^{2}- 2

^{2}) = 30 J.

### Answered by Shiwani Sawant | 23rd Mar, 2020, 11:49: AM

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