Kinetic/Potential Energy of Vehicle on Ramps

[lapo3399]

Homework Statement

Under the influence of gravity g, a vehicle with mass m and initial velocity v0, travels a distance d1 to the bottom of a frictionless ramp ("a" degrees above horizontal). It then begins to travel up a ramp ("b" degrees above horizontal) with friction coefficient u. What distance d2 will the vehicle travel up the second ramp before stopping? (use energy methods)

2. The attempt at a solution
The kinetic energy gained by the vehicle on the first ramp will be dissipated by the negative work of friction and gravity on the second ramp. Thus:

Ek = Eg

To find a general formula for d2, I found expressions for Ek at the bottom of the first ramp and Eg at the top of the second and set them equal to each other (v1 is the speed at the bottom of the first ramp):

v1^2 = v0^2 + 2ad
v1 = sqrt[ v0^2 + 2*d1*g*sin(a) ]

Ek = 1/2m(v1)^2
Ek = 1/2m[v0^2 + 2*d1*g*sin(a)]
Ek = 1/2m(v0)^2 + d1*m*g*sin(a)

(Wf is work performed by friction, Wg by gravity; Ff is force of friction)

Eg = Wf + Wg
Eg = d2(Ff + Fg)
Eg = d2[u*m*g*cos(b) + m*g*sin(b)]
Eg = d2*m*g[u*cos(b) + sin(b)]

Ek = Eg
1/2m(v0)^2 + d1*m*g*sin(a) = d2*m*g[u*cos(b) + sin(b)]
d2 = [1/2(v0)^2 + d1*g*sin(a)] / [g(u*cos(b) + sin(b)]

Thanks for any help!

 

[Doc Al]

Looks OK to me.

 

[lapo3399]

Thanks :)