Question

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?

Answer 1

7362.2 N = 7.3622 kN

Step-by-step 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

Answer 2

The drag force = 3275.80N

Step-by-step 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