Answer:
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
Answer:
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
