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 = 109 m. Later, the car comes to a dip with a radius of curvature rd = 77 m. Assume that the car maintains a constant speed of v = 23 m/s as it goes over the hill and through the dip.a)The actual weight of the driver, as measured on a flat stretch of road, is 535 N. What is the apparent weight of the driver at the top of the hill?b) What is the apparent weight of the driver at the bottom of the dip?

The apparent weight is the force felt by the person, which in these cases is the normal force, or the force that the ground, or in this case, the car, exerts on the person.

Both at the top of the hill and the bottom of the dip the sum of the normal force and the weight must be the centripetal force F, since the stretch there must be a circle. For the hill we will have [tex]N_h-W=-F_h[/tex], where h stands for hill and the centripetal force must be downwards, and for the dip [tex]N_d-W=+F_d[/tex], where d stands for dip and the centripetal force must be upwards.

The equation for centripetal force is [tex]F=m\frac{v^{2}}{r}[/tex], and taking into account that W=mg, where [tex]g=9.8m/s^2[/tex] we have everything we need.

First we calculate [tex]N_h=W-F_h=W-\frac{W}{g}\frac{v^{2}}{r_h}[/tex], which gives 270N, and then [tex]N_d=W+F_d=W+\frac{W}{g}\frac{v^{2}}{r_d}[/tex], which gives 910N.