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?

The drag force = 3275.80N

Explanation:

Using the law of conservation of energy

Initial KE = 1/2mv^2 = 1/2 * 9410 * 30.7^2=4434415.45J

PE = mgh = 9410*9.8*11.9 = 1097394.2J

Car's KE = 4434415.45 - 1097394.2 = 3337021.25J

1/2mv^2 = 3337021.25

1/2*9410*v^2=3337021.25

V^2 = 3337021.25/4705

V=Sqrt (709.25)

V = 26.63m/s

The drag force caused by the velocity to decrease from 26.63m/s to 12.4m/s as it moved 355m is calculated thus:

Finding deceleration using the motion equation

V^2=Vf^2 + 2ad

709.25 = 153.76 + 2*355*a

670a = - 555.49

a = - 0.829m/s^2

F = ma = 9410 * (-0.839) = - 7800.9N

Fp = 9410 * 9.8*11.9/355

Fp = 3275N

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