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LinearBeat
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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 6.0° and the coefficient of friction is 0.46. Use work and energy to find the length of a ramp that will stop a 10,000-kg truck that enters the ramp at a speed of 35 m/s (75 mph).
F_f=Force of Friction (F_f=μF_n; F_n=Normal Force)
m=Mass of the truck
U=Potential Energy, U-i= initial U, U_f=final U
L=Length of the ramp
K=Kinetic Energy, K_i=initial K, K_f=final K
W=Work (in this case done by friction)
v=initial velocity of the truck
Looking at the diagram I drew I determined:
F_f=μmg/cosθ
U=Lsinθmg
ΔK=ΔU-W
K_f-K_i=U_f-U_i-W
-K_i=U_f-W
-1/2mv^2=Lsinθmg-(μmg/cosθ)L
-1/2v^2=L(sinθg-μg/cosθ)
-v^2/(2(sinθg-μg/cosθ))=L
Using this process I've gotten the wrong answer, but I don't understand what I've done wrong and how I need to revise it.
Thanks!
F_f=Force of Friction (F_f=μF_n; F_n=Normal Force)
m=Mass of the truck
U=Potential Energy, U-i= initial U, U_f=final U
L=Length of the ramp
K=Kinetic Energy, K_i=initial K, K_f=final K
W=Work (in this case done by friction)
v=initial velocity of the truck
Looking at the diagram I drew I determined:
F_f=μmg/cosθ
U=Lsinθmg
ΔK=ΔU-W
K_f-K_i=U_f-U_i-W
-K_i=U_f-W
-1/2mv^2=Lsinθmg-(μmg/cosθ)L
-1/2v^2=L(sinθg-μg/cosθ)
-v^2/(2(sinθg-μg/cosθ))=L
Using this process I've gotten the wrong answer, but I don't understand what I've done wrong and how I need to revise it.
Thanks!