Question

a mass m slides down a fixed plane inclined at an angle α to the horizontal. find the distance that it will move on the horizontal plane after covering the entire length of the inclined plane. the height of the plane is h and the coefficient of friction over both surfaces is μ.

Answer 1

Inclination angle = α
height of plane = h
friction coefficient = μ

Initial potential energy = mgh  (assuming horizontal plane to be of 0 potential energy)
Initial kinetic energy = 0
Final potential energy = 0
Final kinetic energy = 0
Thus change in energy = mgh

The energy is lost due to friction.

Length of inclined plane  = h.cosec(α)
let it travels a distance x on horizontal plane.
Energy lost due to friction = μ[mg.cos(α)(h.cosec(α)) + mg.x]

So mgh = μ[mg.cos(α)(h.cosec(α)) + mg.x]
⇒ mgh = μmg[cos(α)(h.cosec(α)) + x]
⇒h/μ = cos(α)(h.cosec(α)) + x
⇒h/μ = h.cot(α) + x
⇒x = h[1/μ - cot(α)]

Thus it will travel h[1/μ - cot(α)] on horizontal plane.