Kinetic energy and frames of refrence

chuy

Hi!

When we measure the Kinetic energy, which is the correct frame of reference that we must choose?

Thanks

pmb_phy

You may choose any frame you like so long as you also measure all quantities of interest in that frame. I.e. Kinetic energy is an observer defined quantity.

Pete

chuy

Thank you

But I'm a little confused.

For example. A car with mass M=1000kg Initially in rest with respect to the ground accelerates until reaching v=50m/s. Use a fuel that renders R=50 000J/kg (1kg of fuel becomes 50000 jouls)

[tex]E=\Delta K=\frac{1}{2}Mv^2=mR[/tex] (m is the mass of fuel that the car spend)

[tex]m=\frac{Mv^2}{2R}=25kg[/tex]

But if we change the frame of reference (also inertial) the car spends different fuel

What happens?

Triss

changing frame of reference does not chance anything. Your car is initially at rest v_i=0m/s and accelerates to v_f=50km/h=13.89m/s. that gives you the diffrence in kinetic energy \Delta K= 0.5M(v_f-v_i)^2.

If you are in a frame of refrence moving at 5m/s relative to the ground, the car is not initially at rest, but is moving with a speed v_i=-5m/s and accelerates to v_f=8.89m/s. Thus, the diffrence in kinetic energy is the same. (Remember to be consistent with your units, and transform km/h into m/s before you do your calulations - your value for m is wrong)

chuy

Thank you for you answer

Sorry, I will change m/s instead of km/s.

Doc Al

Careful! This reasoning is incorrect; in general:
[tex]\Delta{KE} \ne \frac{1}{2}m (v_f - v_i)^2[/tex]

Instead:
[tex]\Delta{KE} = \frac{1}{2}m (v_f^2 - v_i^2)[/tex]

Yes, the change in KE of the car does depend on the reference frame used; to understand what's going on, you need to consider the change in KE of everything, including the earth itself. Here's a post that might help: http://www.physicsforums.com/showpos...9&postcount=12

chuy

Well, if we change of frame of reference by other that one moves to -5m/s relative to ground. The initial speed of the car in that frame of reference will be v_0=5m/s and the end v_f=55m/s
[tex]E=\Delta K=\frac{1}{2}Mv_f^2-\frac{1}{2}Mv_0^2=\frac{1}{2}M(v_f^2-v_0^2)=\frac{1}{2}M(55^2-5^2)=1.5\times 10^6 J[/tex]

[tex]m_{fuel}=\frac{1.5\times 10^6 J}{R}=30kg[/tex]

It spent different fuel.

Kind regards

Edit:
Thanks Doc Al. I see that.