Question

A 9410-kg car is travelling at 26.8 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 3.90 × 102 m along the exit ramp the car\'s speed is 13.1 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

Consider the case when there is no drag force, then by the conservation of kinetic and potential energy to determine the car’s velocity at the top of the ramp.

Now -

Initial KE = ½ * 9410 * 26.8^2 = 3379319.2 J
Final PE = 9410 * 9.8 * 14.9 = 1374048.2 J
To determine the car’s kinetic energy at the top of the ramp, subtract these two numbers -

Final KE = 3379319.2 – 1374048.2 = 2005271.0 J

=> ½ * 9410 * v^2 = 2005271
=> v^2 = 2005271 / 4705 = 426.2

=> v = 20.64 m/s

Now, the drag force caused the car’s velocity to decrease from approximately 20.64 m/s to 13.1 m/s as it moved 3.90 x 10^2 = 390 meters. To determine the drag force, we need to determine the deceleration. Use the following equation.

vf^2 = vi^2 + 2 * a * d
426.2 = 13.1^2 + 2 * a * 390

=> 780 * a = 426.2 - 171.61

=> a = 0.33 m/s^2

Therefore, the drag force exerted on the car, F = m * a = 9410 * 0.33 = 3105.3 N