Stopping 15000 kg Truck with Friction and Gravity

AI Thread Summary
The discussion focuses on calculating the length of a gravel ramp needed to stop a 15,000 kg truck entering at 35 m/s, considering friction and gravity. The truck's kinetic energy is equated to the work done by friction and gravity. The frictional force is determined using the coefficient of friction (0.40) and the gravitational force on the incline. The calculations involve using trigonometry to relate the height of the ramp to its length, but the user encounters issues with the math, leading to incorrect results. The final calculated length of the ramp is 12.4 meters, but the user is uncertain about the accuracy of this solution.
marissa12
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Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of friction in the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6 deg. and the coefficient of friction is 0.40.

Use work and energy to find the length of a ramp that will stop a 15,000 kg truck that enters the ramp at 35m/s

Delta(kinetic energy)=W(net)

W(friction)= -u_k*m*g*d(distance)
W(gravity)= -m*g*(delta y)

0.5*m*v^2= W(friction) and then you solve for d.. is this right? because i got some really weird answers
 
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Find y in terms of d using trig and then do W(Friction) + W(gravity) = KE
 
maybe i screwed up the math..because that didnt work

0.5*m*v^2=(-u_k*m*g*D)-(m*g*D*sin(6))
9187500=-58860d-146344d
9187500=-205204d
d=12.4

and that was still wrong, and i can't see where i messed up in the math
 
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