How much work would it take to horizontally accelerate an object with a mass of #2"kg"# to #9"m/s"# on a surface with a kinetic friction coefficient of #3#?

1 Answer
Feb 4, 2016

It can be anything greater than #81"J"#.


From the Work-Energy Theorem, the work done equals to the change in kinetic energy of the object and the energy loss due to friction.

The change in kinetic energy, #"KE"#, is given by

#Delta "KE" = 1/2 (2"kg") [(9"m/s")^2-(0"m/s")^2] = 81"J"#.

It will be "81"J"# no matter how you push the object, so the work done will be at least that.

The work done against friction on the other hand, can vary from infinitesimally small to infinitely big. Let's consider the 2 extreme cases intuitively.

The fundamental reason is that the energy lost is the product of the #"kinetic frictional force" = mu m g# (a finite constant) and the distance traveled.

In the case where a huge force is applied, the objects reaches the final speed of #9"m/s"# in almost no time at all. In such a small amount of time, the distance traveled is also approximately zero. Therefore frictional loss is minimal.

On the other end of the spectrum, consider applying a constant force of #mu m g#. Since it is equal in magnitude as the opposing friction force, the net force is zero and the object will not speed up. But as it is moving, energy is constantly being lost by friction. This treatment can be extended for arbitrarily long and you can lose as much energy as you want in this way.