Truck brakes can fail if they get too hot. In some mountainous areas, ramps of l

Question

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 rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0 / circ and the coefficient of rolling friction is 0.44. Use work and energy to find the length of a ramp that will stop a 15,000 /rmkg truck that enters the ramp at 30/rmm/s .

Explanation / Answer

The change in KE always equals the total work done (by all forces) on the object. In our case, we want the object to go from speed "v" (35m/s) to zero, so the change in KE will be: ?KE = -½mv² Two forces do work on the truck: Force #1: Gravity. It does this much work on the truck: Work_grav = -(weight)×(change in height) Work_grav = -mg(Lsin?) [EDIT: "Pointy" forgot the "sin?" in this term!] (where "m" = truck's mass; "L" = length of slope (not yet known); "?" = 6 degrees). Note: Work_grav is negative because the force is in a direction opposing the motion. Force #2: Friction. It does this much work on the truck: Work_fr = -(friction force)×(length of slope) Work_fr = -((Normal force) × µ) × L Work_fr = -mg(cos?)µL (where "µ" = coefficient of friction (0.40)) Again, Work_fr is negative because the force is in a direction opposing the motion. Put it all together: ?KE = (total work) -½mv² = Work_grav + Work_fr -½mv² = -mg(Lsin?) + -mg(cos?)µL The rest is just algebra: just solve for "L". (Interesting note: the "m" cancels out completely: so it doesn't even matter what the mass of the truck is!)

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