Answer:
4899.47 N
Explanation:
mass of the car = 7780 kg, u, initial speed = 30.7 m/s, v, final speed = 13.3 m/s
d = 404 m and h = 13.1 m
initial kinetic energy of the car = potential energy attained by the car climbing the ramp + the work done by friction against the motion + final kinetic energy
1/2 mu² = mgh + (F × d) + 1/2mv²
3666286.1 J = 998796.4 J + ( 404 F) + 688102.1 J
404 F = 3666286.1 J - 998796.4 J - 688102.1 J
F = 1979387.6 / 404 = 4899.47 N
We have that for the Question, it can be said that the average drag force F drag exerted on the car
F=5239.18N
From the question we are told
A 7780 ‑kg car is travelling at 30.7 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 404 m along the exit ramp, the car's speed is 13.3 m/s, and it is h = 13.1 m above the freeway. What is the magnitude of the average drag force F drag exerted on the car?
Generally the equation for the work energy theorem is mathematically given as
[tex]K.E_1-F*d=K.E_2+mgh\\\\Therefore\\\\1/2* m* v_0^2 - F * d = 1/2 * m * v^2 + mg h\\\\0.5 * 7780* 30.7^2 - F * 404 = 1/2 * 7780* 13.3^2 + 7780 * 9.8 * 11.3\\\\-404F=( 1/2 * 7780* 13.3^2 + 7780 * 9.8 * 11.3)-(0.5 * 7780* 30.7^2)\\\\F=\frac{-2116626.8}{-404}\\\\[/tex]
F=5239.18N
Therefore
the average drag force F drag exerted on the car
F=5239.18N
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