Question

A 7780-kg car is travelling at 29.7 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 397 m along the exit ramp the car\'s speed is 12.0 m/s, and it is h = 14.9 m above the freeway. What is the magnitude of the average drag force exerted on the car?

Answer 1

I'm assuming the driver coasts all the way along the ramp.

So from the conservation of energy we have KE = DE + PE + ke; total kinetic energy starting up the ramp, drag energy DE = ? is what you're looking for, PE = mgh is the stored energy when h = 14.9 m up the ramp, and ke = 1/2 mv^2 is the remaining kinetic energy when v = 12.0 m/s. KE = 1/2 mV^2 when V = 29.7 m/s.

Solve for DE = KE - PE - ke = m(1/2 V^2 - 1/2 v^2 - gh) = 7780*(29.7^2/2 - 12.0^2/2 - 9.8*14.9) = 1735134.5 J drag energy.
S = 397 m coasting distance
So the average drag force D = DE/S = 1735134.5/397 = 4370.6 N.