Solving the Work Energy Theorem for an Observer Moving Left

[vu10758]

A 2kg block at rest on a frictionless surface is pulled for 2s by a 10N horizontal force (pull to the right).

Use the Work Energy Theorem to determine the final velocity from the perspective of an observer moving to the left. Does the WET still work that observer?

I did the calculations and got 14 m/s for the first part. However, I am stuck on the second part. Why does the Work Energy Theorem works for an observer who is moving at constant velocity?

 

[Locrian]

An observer moving to the left would have the effect of giving the block some additional initial and final velocity. Add this into the problem and see how this does or does not affect the answer.

 

[vu10758]

I plugged it in the formula w = (1/2)mv^2 - (1/2)mv_o ^2

w = (1/2)m (v^2-v_o^2)
w = (1/2)(2)(14^2-4^2)
w = 180 J.

This is different from

w = (1/2)(2) (10^2 - 0)
w = 100J

So the work in both cases are different. The answer key tells me that WET still works on the observer. But how? I don't know if I am misunderstanding something but if the work are different, how can WET still work?

 

[OlderDan]

So the work in both cases are different. The answer key tells me that WET still works on the observer. But how? I don't know if I am misunderstanding something but if the work are different, how can WET still work?

Something else besides the velocities is different to the two observers. Think about the definition of work.

 

[vu10758]

The distance is also different. Thanks, I understand it now.