# Finding the energy lost due to non-conservative forces

I stomped across this question and would very much appreciate any form of clarification.

A otter 75kg slides down a hill starting from rest. Hyp = 8.8, height = 6.5, final speed of otter = 9.2 m/s. And it wants me to find how much energy was lost due to non-conservative forces on the hill?

I guess my question is what the term non-conservative means in this context. I am not asking anyone to do this for me, I merely ask for some advice or even which formulas I could use to solve this. Is friction a non-conservative force? If so, do I just have to find how much work was done by friction?

This is literally what I get when I Google "non-conservative force".

From the very first link (to Wikipedia), here's a more detailed explanation:

A conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path. Equivalently, if a particle travels in a closed loop, the net work done (the sum of the force acting along the path multiplied by the distance travelled) by a conservative force is zero.

Which would imply that the work done against a non-conservative force on a closed path is not zero.

By conservation of energy, the amount of energy (potential + kinetic) lost by your otter should be equal to the work done against any non-conservative forces.

• I see, so its basically the gravitational potential energy minus the kinetic energy (taking into acct friction). Would this logic be correct ? Dec 8, 2014 at 8:49
• Not exactly. It is the loss in total mechanical energy. In this case, since it starts from rest, initial kinetic energy is zero, but in general, the loss in energy is $$(PE + KE)_{initial} - (PE + KE)_{final}$$
– pho
Dec 8, 2014 at 8:52
• Ah yes, mechanical energy not just potential or kinetic. That helped a lot thank you. Wish i could upvote it but Im not popular enough. Dec 8, 2014 at 9:00

As a first estimate you can evaluate energy deficit in conversion from initial potential energy to a final kinetic energy: $$\Delta E_{~lost} = mgh_0 - \frac {m{v^{~2}_f}}{2}$$