Homework Help: Better intuition of what it is I am finding.

1. Oct 22, 2011

Sefrez

This is another simple problem that I have probably over thought on and now have some questions.

I am given that the initial velocity of some mass m's vector is v_i = x1 i + y1 j and that its final velocity vector is v_f = x2 i + y2 j. I am then asked to find the net work done on the object that accounted for the change in velocity.

This is simply done by W = ΔK = 0.5m|v_f|^2 - 0.5m|v_i|^2.

I got to thinking though that this must be relative. That is, if you find the change in velocity which would be Δv = v_f - v_i = (x2 - x1) i + (y2 - y1) j. This could then be thought of as a reference frame that observes the final velocity as Δv. In other words, in this reference frame, the initial velocity is zero. That means that the work needed to change its velocity is 0.5m|Δv|^2. This of course is not the same as the work needed from the other reference frame.

Also note, that if at the reference frame that defines the two original velocities (initial, final) where the final has a decrease in velocity, work is negative due to the fact that the force and displacement vectors are in opposite directions.

But going back to the reference frame where initial velocity is 0, applying the same force, the object does move in the direction of the force - because the object is essentially at rest in the current reference frame.

So: Does this mean that in the first calculation of work - this is relative to the frame of reference that initially defined the two velocities? And for the work calculated from the change in the velocity is relative to the reference frame that observes initial velocity as zero?

And if those two are correct, this means then that, work is relative, correct? Maybe this is obvious, maybe not. Possibly even incorrect. Any talk is appreciated.

2. Oct 22, 2011

SammyS

Staff Emeritus
ΔK is not simply 0.5m|Δv|2 if you do the algebra correctly. |Δv|2≠Δ(v2)
ΔK = (1/2)m[Δ(v2)]
= (1/2)m[(vf)2-(vi)2]

= (1/2)m(vf+vi)(vf-vi)​

Now, check to see whether or not this is invariant for various inertial frames of reference.

(Actually, it may not be invariant, but at least you'll be working with the correct quantity.)

Last edited: Oct 22, 2011