Kinetic/Potential Energy of Vehicle on Ramps

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SUMMARY

The discussion focuses on calculating the distance a vehicle travels up a second ramp after descending a first ramp, using energy methods. The key formula derived is d2 = [1/2(v0)^2 + d1*g*sin(a)] / [g(u*cos(b) + sin(b))], where d2 is the distance traveled up the second ramp, v0 is the initial velocity, d1 is the distance traveled down the first ramp, a is the angle of the first ramp, b is the angle of the second ramp, and u is the friction coefficient. The kinetic energy gained on the first ramp is equated to the gravitational and frictional work done on the second ramp to derive this formula.

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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!
 
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