A car travels down a straight country road that leads over hills and through val

Question

A car travels down a straight country road that leads over hills and through valleys. On one particular stretch of road, the car encounters a hill that can be approximated as the top of a circle with a radius rh = 105 m. Later, the car comes to a dip with a radius of curvature rd = 81 m. Assume that the car maintains a constant speed of v = 21 m/s as it goes over the hill and through the dip. The actual weight of the driver, as measured on a flat stretch of road, is 545 N.

What is the apparent weight of the driver at the top of the hill? What is the apparent weight of the driver at the bottom of the dip?

Explanation / Answer

weight of driver = 545 N

mass of driver = m = 545/9.81 = 55.56 kg

V = 21 m/s

R = top hill = 105 m

R = concave dip = 81 m

Fc = centripetal force = mV^2/R = 55.56(21)^2/R = 24502/R

top of hill Fc = 24502/105 = 233.35 N

bottom dip Fc = 24502/81 = 302.5 N

At the TOP of hill apparent weight of driver = normal force = Fc - mg = 311.65 N

At the concave dip apparent weight of driver = normal force = Fc + mg = 847.5 N

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