Determining length of highway deceleration ramp, with friction

Satchmo

1. The problem statement, all variables and given/known data
"At the base of a hill (blah blah) there is a ramp inclined with an angle of 30 degrees and has a surface with a coefficient of friction of 1.5. How long should this ramp be to be able to stop a 80,000lb truck traveling at 80 mph"


2. Relevant equations
(delta)K + (delta)U = -fk * d (K is kinetic energy, U is potential energy, fk is friction force)

fk = u*n (u is frictional coefficient)




3. The attempt at a solution

set Ki. = Uf + fk*d (having fk be positive, since you'd be subracting a negative anyway, right?) Ki = initial K.E., Uf = final P.E.)

(1/2) m v^2 = m g h + u n d (u is frict. coefficient, n is normal force, d is distance or hypotenuse of triangle, h is height.)

for the normal force I got n = m g cos(30)
and h = d sin(30)
{both of the above just using trig}

so (1/2) m v^2 = m g d sin(30) + u m g cos(30) d
m's cancel out
(1/2) v^2 = g d sin(30) + u g d cos(30)
solve for d
d = (v^2) / [g(sin(30) + u cos(30))]

plugging in v = 35.8 m/s (80 mph)
g = 9.8 m/s^2
u = 1.5

gives 72.7 meters. I don't have much faith in this answer. Can I do Ki. = Uf + fk*d for this problem?

Bill Foster

Since it's giving you the coefficient of friction, I think that you'd want to use this equation to solve the problem:

[tex]2a(x-x_0)=v^2-v_0^2[/tex]

Solve for [tex]x-x_0[/tex]

Figure out [tex]a[/tex]

[tex]a=-g(\sin{\theta}+\mu \cos{\theta})[/tex]

Substitute:

[tex]x=\frac{v^2-v_0^2}{2a}=\frac{v_0^2}{2 g(\sin{\theta}+\mu \cos{\theta})}[/tex]