Can Any1 tell me why kienetic energy is equal to both m.v[SUP]2[/SUP] and 1/2

[Arslan]

Can Any1 tell me why kienetic energy is equal to both m.v² and 1/2 m.v²


E=W=F.d
=m.a.d ......=mgh
=m.(v/t).d
=m.(d/t^2).d
=m.d^2/t^2
=m.v^2
So hence proved K.E= m.v^2



But 2.a.s=Vf²-Vi² Newton equatio of motion
2.F/m.s=V²......if initial velocity is zero
F.s=1/2 mv²
W=E=1/2 mv²
So hence proved K.E=1/2 m.v^2

 

[Dale]

Please don't double post

=m.a.d
=m.(v/t).d

a=\frac{dv}{dt}\ne\frac{v}{t}

 

[jtbell]

E=W=F.d
=m.a.d ......=mgh
=m.(v/t).d

In the equation above, v is the final velocity at time t.

=m.(d/t^2).d

In this step you use v=d/t, but the object isn't moving at constant speed v. It starts with speed 0, and increases speed linearly until the end (v). d/t is the average speed which is v/2 for constant acceleration from v=0 to v=v. So v=2d/t.

 

[Arslan]

Thanks i got it

i have still one more question related to energy

If a snooker player hits a ball on frictionless surface...he hits with force 6N and ball starts to move with constant velocity say 10m/s and never stops,
how much work is done by the player and how how much energy the ball has.

If we use W=integral(F.dx)
but F=6 at instant t=0 and F=0 for all other values of t
and d=0 at instant t=0 and d is increasing for all other values of t
So the product remains 0

It mean that work done is zero but energy is transferred to ball i.e. 1/2 mv^2

 

[Doc Al]

If we use W=integral(F.dx)
but F=6 at instant t=0 and F=0 for all other values of t
and d=0 at instant t=0 and d is increasing for all other values of t
So the product remains 0

It mean that work done is zero but energy is transferred to ball i.e. 1/2 mv^2

No, it means that your model is flawed. You can't exert a force for zero time. ∫Fdt & ∫Fdx will both be greater than zero.

 

[D H]

No, it means that your model is flawed. You can't exert a force for zero time.

Well you can conceptually exert a finite force for zero duration. It won't do anything, however.

You can also apply a rather large but finite force F(t) for a very small period of time δt. Taking ∫Fdt to be the change in momentum Δp and letting δt→0 so that Δp remains constant leads to the idea of impulsive forces. These can yield quite good estimates of the behavior if

• You don't care about the deformations and such during that time interval δt during which the force is non-zero
• The changes to the integrated state (position) during that time interval δt are very small
• That large but finite force F(t) overwhelms all other forces during that time interval.

 

[Doc Al]

Well you can conceptually exert a finite force for zero duration. It won't do anything, however.

True.