QUESTION 1: For a 1,582-kg car with a drag constant D=0.49 kg/m, which is moving
ID: 1281312 • Letter: Q
Question
QUESTION 1: For a 1,582-kg car with a drag constant D=0.49 kg/m, which is moving at a speed of 39 mph, what is the magnitude of the maximum acceleration (right at this speed) at which the car can further speed up. The coeffcicient of kinetic friction is 0.55 and the coefficient of static friction is 0.79.
Calculate the answer to at least 3 significant figures (this is too accurate for practical purposes, but will test whether you really understand this problem.)(Note, this will be the limit due to friction; the real maximum might be further limited by engine power or torque.)
QUESTION 2: For a 1,478-kg car with a drag constant D=0.46 kg/m, which is moving at a speed of 35 mph, what is the magnitude of the maximum acceleration (right at this speed) at which the car can be slowed down. The coeffcicient of kinetic friction is 0.62 and the coefficient of static friction is 0.75.
Calculate the answer to at least 3 significant figures (this is too accurate for practical purposes, but will test whether you really understand this problem.)(Note, this will be the limit due to friction; the real maximum might be further limited by engine power or torque.)
QUESTION 3: Find the terminal velocity for a 90-kg skydiver. Assume a constant of D=0.25 kg/m. Convert your answer to miles per hour and use mph as units.
QUESTION 4: Find the constant D for a 81-kg skydiver with open parachute, such that the terminal velocity is equivalent to that of the skydiver jumbing from a height of 1.5 m without a parachute.
QUESTION 5: What is the magnitude of the maximum possible acceleration for a car, when the coefficient of static friction is 0.8 and the coefficient of kinetic friction is 0.68? Ingore air drag.
Explanation / Answer
39 mph = 17.43 m/s
Fd = D v2
frictional force = muk m g
net force = D v2 + (muk m g)
ma = D v2 + (muk m g)
a = D v2 / m + muk g
= [0.49 / 1582] 17.432 + 0.55 * 9.8
= 5.48 m/s2
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35 mph = 15.65 m/s
Fd = D v2
frictional force = muk m g
net force = D v2 + (muk m g)
ma = D v2 + (muk m g)
a = D v2 / m + muk g
= [0.46 / 1478] 15.652 + 0.62 * 9.8
= 6.15 m/s2
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4) V = sqrt [2 * 1.5 * 9.81]
= 5.42 m/s
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5) a = ug
= 0.68 * 9.8
a = 6.67 m/s2