1. Figure 1 shows a car going over the crest of a hill. The hill has a circular

Question

1. Figure 1 shows a car going over the crest of a hill. The hill has a circular cross section (at least at the top) with a radius of 40.0 m as shown in the gure. At the moment it goes over the top of the hill the car is going 15.0 m/s. However, the driver has applied the brakes. The tires are not slipping. The coecient of static friction between the road and the tires is 0.900 and the coecient of kinetic friction is 0.750. (a) The driver is sitting on a scale (don't you always drive sitting on a scale??). The driver has a mass of 60.0 kg. What does the scale read as the car is going over the top of the hill? (b) What is the maximum rate at which the car can slow down? (c) Suppose the driver applies the brakes too hard and the tires start to slip (i.e. the car skids). What will be the rate at which the car slows down now? (d) What is the maximum speed that a car can over this hill without its tires leaving the road? Additional questions for practice (will not be marked ) Knight: CQ6-2, CQ6-3, CQ6-7, CQ6-10, CQ6-11, CQ6-15, P6-7, P6-27, P6-31, P6-43, P6-55, CQ8-2, CQ8-5, P8-21, P8-23, P8-37, P-8-43

Explanation / Answer

a) sum forces in the y with down being positive

mg - N = mv^2/r

N = m g - mv^2/r = 60*9.81 - 60*15^2/40= 251.1 N

so reads 251.1/9.81= 25.6 kg

b) max rate is when Fstatic = Fstatic max

F = u N = m a

a = 0.9*251.1/60= 3.77 m/s^2

c) now since slipping use kinetic

a = 0.75*251.1/60 = 3.14 m/s^2

d)

at max speed N = 0

m g = mv^2/r

v = sqrt( gr) = sqrt(9.81*40)= 19.8 m/s

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