Please show full wok and correct answer. Best answer gets 1,700 points. An 4000

Question

Please show full wok and correct answer.

Best answer gets 1,700 points.

An 4000 lb car on a level road draws 8.00 hp from its engine in order to drive at a constant 30.0 mph. (a) Calculate the effective drag force on the car. Assume this drag force is present. when the car travels at 30.0 mph (i.e., use it for the rest of the problem). (b) If instead the car encounters a 10% grade (the road rises vertically 1 ft for every 10 ft the car travels along the road), what power (in hp) is required from the engine to maintain a steady 30.0 mph? (c) What power (in hp) must the engine provide to drive down a 1.00% grade? (d) Determine the grade down which this car will effortlessly coast at 30.0 mph. You may find the following useful: 30 mph = 44 ft/s; 1 hp = 550 ft lb/s.

Explanation / Answer

a) Since the car is moving at a constant velocity, the net force on the car is zero.

Therfore, the power derived from the car = power consumed by the drag force

or, Fdrag x v = Power supplied

Power supplied = 8 x 746 W = 5968 W

v = 30 mph = 13.41 m/s

Fdrag =  5968 / 13.41 = 445.04 N

b) When going up a slope at a constant velocity,

Power supplied = Power used by drag force + Power used by force of gravity

Or, Power supplied = 5968 W + 1814.369 x 10 x 0.1 W = 7782.369 W

c) When going down a slope at a constant velocity,

Power supplied + Power provided by force of gravity = Power used by drag force

or, Power supplied = 5968 W -  1814.369 x 10 x 0.01 W = 5786.56 W

d) For the car to go down effortlessly,

Power provided by force of gravity = Power used by drag force

1814.369 x 10 x grade W = 5786.56 W

or, grade = 0.32 = 32 % gradiant

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