Let's imagine there's a block of mass m sitting on a rough surface with a kinetic friction coefficient $\mu$. It's being pulled with a constant force $A$ at $h$ degrees above the horizontal and is displaced a distance $d$. The work done by $A$ on the block is positive and is the horizontal component of $A$ times the displacement:
$$A\cos (h)*d.$$
I thought this meant that the work done by friction would be either the negative of that amount or else
$$\mu*(mg-A\sin (h))*d$$
but apparently not?
Best Answer
You were very close. The work done by the constant force of kinetic friction is W_fric = Fdcos(a) where a is the angle between the friction and the displacement. Kinetic friction always points in the direction opposite the motion, so a equals 180°. This was your error. If the force and displacement point in opposite directions the angle bewteen them is 180° not 0°.
F = un = u(mg-Asin(h)) (as you have). d=d (highly insightful). And cos(a) = cos(180°) = -1. So W_fric = -u*(mg-Asin(h))*d, which is really just the negative of your answer. Also as a general rule, kinetic friction always points opposite to the direction of motion and hence always does negative work. Thanks for the question, I hoped my answer helped you out, and have a nice day.