inclined plane revisited

[tricky_tick]

Movers want to set the ramp of their truck so that the work they do against the combination of gravity and friction is a minimum for crates moving up the ramp with constant velocity. µ is the coefficient of kinetic friction and θ is the angle between the ramp and the ground. For the work to be a minimum, they must choose:

a. tan θ = µ

b. tan θ = -µ

c. tan θ = -1/µ

d. tan θ = 1/µ

e. tan θ = 1 - µ

The obvious solution of setting the derivative of work equal to zero finds the maximum work. The problem asks for the minimum?

[HallsofIvy]

The word done is "Force*distance".

The force necessary to move an object (I'm assuming they are sliding it) is the component of weight along the ramp: mg sin(θ) (draw a picture and look at the right triangles!) plus the friction force: μ times the component of weight perpendicular to the ramp, mg cos(θ).
Force= mg sin(θ)+ mg μ cos(θ).

The distance is the length of the hypotenuse of the right triangle formed by the ramp: h/cos(θ) where h is the height of the ramp (which I am assuming is fixed: the height of the truck bed).

The work done is hmg(sin(θ)+ μcos(θ)/cos(θ).

What value of θ minimizes that?