Question

An object, previously at rest, slides `5 m` down a ramp, with an incline of `(3pi)/8`, and then slides horizontally on the floor for another `2 m`. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Answer 1

Start with a force diagram for the object both on the incline and on the straight section.

Considering the incline section, and applying Newton's 2nd Law of motion in the direction parallel to the incline, together with the definition of friction, we get :

`sumF=ma`

`thereforemgsintheta-mu_kmgcostheta=ma`

`therefore a=g(sintheat-mu_kcostheta)...............1`

Now using energy considerations on the inclined section, we note that potential energy at the top less work done by friction in sliding to the bottom equals left over kinetic energy at the bottom.

`therefore mgh-f_kx=1/2mv^2`

`therefore mg(5sintheta)-5mu_kmgcostheta=1/2mv^2`

`therefore5g(sintheta-mu_kcostheta)=1/2v^2................2`

Now using energy considerations along the flat section, application of the Work-Energy Theorem yields :

`W_f=DeltaE_k`

`thereforemu_kmg*2=1/2mv^2`

`therefore 2mu_kg=1/2v^2....................3`

Now comparing equations 2 and 3 we observe that they are equal and hence

`2mu_kg=5g(sintheta-mu_kcostheta)`

`therefore mu_k=(5sintheta)/(2+5costheta)`

`=(5sin(3pi/8))/(2+5cos(3pi/8))`

`=1.18`

Since this value of `mu_k !in [0,1]`, it implies the question is unrealistic and does not happen in practice as proposed in the question.
(Unless I made a mistake somewhere in my calculations - I request other contributors to please check for me).