Question

# A large tractor-trailer of total mass M = 10 metric tons (10,000 kg) is traveling down...

A large tractor-trailer of total mass M = 10 metric tons (10,000 kg) is traveling down I-70 in Colorado, coming down from the Rocky Mountains into Denver. A segment of this road that is 25 kilometers long descends a total of 1000 meters. Suppose that the truck driver wants to drive at a constant speed of v = 50 km/hr.

(a) What is the work done on the truck by its brakes during this motion? (Give both the sign and the magnitude.) Does this depend on how fast the truck drives?

(b) What is the power applied to the truck by its brakes during this motion? Does this depend on how fast the truck drives?

(c) The coefficient of friction of the metal that comprises brake pads decreases as they are heated. If they get too hot, they will not be able to apply the required braking force. Explain why trucks must drive very slowly down a long hill such as this one, but do not need to be so cautious on a short hill.

Your answer to this problem should include mostly words, rather than mathemat- ics; you do not need to do any calculations here. However, two things that may be helpful to you:

If an amount of heat energy Q is added or removed to an object, its temper- ature will increase or decrease. This amount is proportional to the amount of heat added or removed; as an example it takes 4000 J of energy to raise the temperature of a liter of water by 1 degree, so 8000 J will raise it by two degrees, and so forth.

Once the brakes heat up, there is some rate Pd (“power dissipated”) at which the brakes will cool.

(Picture)

A warning sign on I-70 approaching Denver, shortly after the Eisenhower Tunnel, over 11,000 feet (3400 meters) elevation. This is the highest interstate highway in the USA.

a)

Sin = height /length = 1000/25000 = 1/25

for constant velocity , parallel to incline , force equation must be

fk = mg Sin

fk = 10000 x 9.8 x (1/25) = 3920 N

L = length of incline = 25000

work done by the brakes = work done frictional force = fk L Cos180 = - 3920 x 25000 = - 9.8 x 107 J

b)

t = time taken to travel = length of incline / speed = 25000/13.88 = 1801.2 sec

Power is given as

P = W/t = (- 9.8 x 107 )/1801.2 = - 5.4 x 104 Watt

c)

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