A 9410 9410 ‑kg car is travelling at 30.7 30.7 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 355 355 m along the exit ramp, the car's speed is 12.4 12.4 m/s, and it is h = 11.9 h=11.9 m above the freeway. What is the magnitude of the average drag force F drag Fdrag exerted on the car?

7362.2 N = 7.3622 kN

Explanation:

From conservation of energy principle

The kinetic energy change of the car = the workdone by the weight + workdone by drag force.

Let v₁ = 30.7 m/s and v₂ = 12.4 m/s be the initial and final velocities of the car,

So 1/2mv₂² - 1/2mv₁² = mgh + Fd where m = mass of car = 9410 kg, h = height of ramp = 11.9 m, g = 9.8 m/s², F = drag force and d = distance moved by drag force = 355 m

So, 1/2 × 9410 × 12.4² - 1/2 × 9410 × 30.7² = 9410 × 9.8 × 11.9 + F × 355

(723440.8 - 4434415.45) J = 1097394.2 J + 355F

3710974.65 J = 1097394.2 J + 355F

(3710974.65 J - 1097394.2 J)= 355F

-2613580.45 J = 355F

F = -2613580.45/355 = -7362.2 N

So the magnitude of the drag force is 7362.2 N = 7.3622 kN

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