How does the angle of an inclined plane relate to its efficiency, given the coefficient of friction between a body and the plane?

Let’s define the efficiency first. The efficiency is basically the ratio between what we want over what we pay for. So, we want to lift an object up to a height, say H. In the ideal scenario we lift that object on the vertical, consuming a work equal to mgH. This is the ideal scenario, this is what we want, but we must use an inclined plane for lifting it to the desired height, therefore we will have to do more work, as we have to overcome the friction on the plane. In order to lift the object, we must apply a force F, parallel to the plane, where F is given by: F=Mumg*cos(alpha) + m * sin(alpha).

Thus, the work done = F * D, where D is the distance travelled along the plane to reach the desired height H.

Hence, D = H / sin(alpha)

Efficiency = (mgH)/ ((Mumgcos(alpha) + mg*sin(alpha)) *H/ sin(alpha))

Simplifying the fraction gives:

 Efficiency = 1/(1+Mu*cot (alpha)) - This is the relation required. 

Notations used:

m = mass of the object 

g = gravitational constant

alpha = the angle of the inclined plane 

Mu = coefficient of friction between the body and the inclined plane.