State kinetic energy and derive an expression for it
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Let a body of mass ‘m’ is moving with a velocity ‘v’.A force ‘F’ is applied on it to stop its motion for which the retardation ‘h’ is produce in the body and the body covers a distance ‘s’ before coming rest.
Work done by the force to stop the motion of the body is the measure of kinetic energy .
Here
Initial velocity = v
Final velocity = 0(zero)
From Newton’s 2nd law
F = ma
or, a = F/m
From 3rd equation of motion
0=v^2 - 2as
or, 2as = v^2
or, 2 × F/m × s = v^2
or, s = (mv^2) /2F
Therefore , workdone by the force,
W = Fs
or, W = F ( mv^2 )/2F
or, W = 1/2 mv^2
Hence,
Kinetic energy = 1/2mv^2.
Work done by the force to stop the motion of the body is the measure of kinetic energy .
Here
Initial velocity = v
Final velocity = 0(zero)
From Newton’s 2nd law
F = ma
or, a = F/m
From 3rd equation of motion
0=v^2 - 2as
or, 2as = v^2
or, 2 × F/m × s = v^2
or, s = (mv^2) /2F
Therefore , workdone by the force,
W = Fs
or, W = F ( mv^2 )/2F
or, W = 1/2 mv^2
Hence,
Kinetic energy = 1/2mv^2.
Answered by
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the energy possessed by a body by virtue of its state of motion is callew kinetic energy..
kinetic enrgy= 1/2*mass*velocity^2.
k= p^2/2m
where p is momentum.
kinetic enrgy= 1/2*mass*velocity^2.
k= p^2/2m
where p is momentum.
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