Why Does Work Done on an Object Differ Between Inertial Frames?

[chingcx]

I've 2 questions:

First, is there any particular reasons why accelerating an object from 1m/s to 2m/s requires less work than accelerating from 2m/s to 3m/s?

Second, when I see another inertial frame moving 1m/s w.r.t. me, in which there is an object moving 1m/s in the same direction w.r.t. that frame. When it is subjected some forces and accelerated to 2m/s in the frame (when it is moving 3m/s in my frame).
Now the work done on the object according to someone in that frame, is not the same as the work done I said. Why is there a contradiction?

Thanks in advance

 

[Hootenanny]

I've 2 questions:

First, is there any particular reasons why accelerating an object from 1m/s to 2m/s requires less work than accelerating from 2m/s to 3m/s?

It follows from the definition of work, try googling for the derivation of kinetic energy.

Second, when I see another inertial frame moving 1m/s w.r.t. me, in which there is an object moving 1m/s in the same direction w.r.t. that frame. When it is subjected some forces and accelerated to 2m/s in the frame (when it is moving 3m/s in my frame).
Now the work done on the object according to someone in that frame, is not the same as the work done I said. Why is there a contradiction?

There isn't a contradiction, you are completely correct. The kinetic energy of an object is dependent on the frame of reference, as is the work done on an object.

 

[shamrock5585]

thats why its is all "relative"

 

[rcgldr]

The frame of reference should be the point of application of force. For example, for a car on a road, the point of application of force is the contact patch with the pavement, and the frame of reference is the surface of the Earth at the point of contact. If the car were to be placed on a very long flatbed truck, then the frame of reference would be the surface of the flatbed truck. In the case of a propellor driven aircraft, the point of application of force is the air itself, which could be moving with respect to the surface of the Earth (headwind or tailwind).

Work done equals force times distance relative to the point of application (or the line integral of F ds from point A to point B).

In the case of a rocket in outerspace, there's nothing to apply a force to, so there's no change in momentum. Instead the rocket expells small bits of itself, spent fuel, at high velocity, and the thrust is the result of the mass of the spent fuel times it's net overall acceleration. Work is done on the spent fuel accelerated in one direction, and on the rocket accelerated in the opposite direction. The result of this work is a change in the total kinetic energy of spent fuel and rocket, and the sum of these two components of energy will be the same regardless of the frame of reference (within reason). The proper frame of reference is the rocket's engine, since that it the point of application of force.