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 = 75 m. Assume that the car maintains a constant speed of v = 23 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 525 N. What is the apparent weight of the driver at the top of the hill?

W_h=........ N

- What is the apparent weight of the driver at the bottom of the dip?

W_d=......... N

Answer 1

weight of driver = 525 N

mass of driver = m = 525/9.81 = 53.5 kg

V = 23 m/s

R = top hill: 105 m & R = concave dip: 75 m

Fc = centripetal force = mV²/R = 53.5(23)²/R = 28301.5/R

top of hill Fc = 28301.5/105 = 269.5 N

bottom dip Fc = 28301.5/75 = 377.4 N

At the TOP of hill apparent weight of driver = normal force = Fc - mg = 525 - 269.5

? 256 N ANS

At the concave dip apparent weight of driver = normal force = Fc + mg =525 + 377.4 ? 902 N ANS