Does my understanding of work, velocity, and friction make sense

[Niko Bellic]

Is this following correct? :

When you push a crate along the floor at a constant velocity for a long time, you get tired which indicates that you are tranferring some of your energy to the object, i.e. you are doing "work" on the object. But according to the definition of work^[1] which equates work to a change in kinetic energy (change in velocity), there is no work done on the crate since its velocity is constant. This is because while you do in fact do work on the crate, the force of friction does equal work in the opposite direction of you, causing the net work done on the crate to be zero.

In the other definition of work where work is a line integral^[2], although you are exerting a force on the crate over a distance which results in a positive value of W, the force of friction is equal and opposite, resulting in two integrals that void each other.

[1] http://upload.wikimedia.org/math/b/f/2/bf240d906ff97b33fc3e60f2508ab671.png
[2] http://upload.wikimedia.org/math/7/6/8/7680d79cfc1c61f21fe00e1089a9493b.png

THANKS!

 

[cepheid]

Is this following correct? :

When you push a crate along the floor at a constant velocity for a long time, you get tired which indicates that you are tranferring some of your energy to the object, i.e. you are doing "work" on the object. But according to the definition of work^[1] which equates work to a change in kinetic energy (change in velocity), there is no work done on the crate since its velocity is constant. This is because while you do in fact do work on the crate, the force of friction does equal work in the opposite direction of you, causing the net work done on the crate to be zero.

In the other definition of work where work is a line integral^[2], although you are exerting a force on the crate over a distance which results in a positive value of W, the force of friction is equal and opposite, resulting in two integrals that void each other.

[1] http://upload.wikimedia.org/math/b/f/2/bf240d906ff97b33fc3e60f2508ab671.png
[2] http://upload.wikimedia.org/math/7/6/8/7680d79cfc1c61f21fe00e1089a9493b.png

THANKS!

This all sounds fine. Another way to think about it is that to compute the work done on an object, you must consider the net force on the object, which is zero in the example you are considering. But yes, in such an instance, the work done by all the individual forces acting on the object must sum to zero.

Note that equation [2] is the definition of work, whereas equation [1] is a theorem about work which can be derived starting from that definition. This theorem is called the work-energy theorem. I guess my point is that the work-energy theorem not a definition of work, but rather a result that follows from the definition.

 

[Bhargava2011]

Your understanding of the concepts appears to be absolutely fine.
When you apply a force to the moving object, you are performing work on the object so, its kinetic energy must increase but, at the same time frictional force is doing a negative work on the object or extracting the energy of the object. The power given by you to the object is opposite and equal to the power of the frictional force. And therefore the velocity of the object remains same and the energy dissipated by you appears as heat energy due to friction.
Thanks.