Question

A 8.10×10^3 ‑kg car is travelling at 25.8 m/s when the driver decides to exit the freeway by going up a ramp. After coasting 3.90×10^2 m along the exit ramp, the car's speed is 13.1 m/s, and it is ℎ=12.5 m above the freeway. What is the magnitude of the average drag force ????drag exerted on the car?

Answer 1

Answer:

The magnitude of the average drag force is 2412.34 N.

Explanation:

Given that,

Mass of car $m=8.10\times10^{-3}\ kg$

Velocity v = 25.8 m/s

Distance $d= 3.90\times10^{2}$

Speed of car = 13.1 m/s

Height = 12.5 m

We need to calculate the magnitude of the average drag force

Using equation kinetic energy

$K.E_{i}=K.E_{f}+P.E+F_{d}$

$\dfrac{1}{2}mv_{i}^2=\dfrac{}{}mv_{f}^2+mgh+F\times d$

Where, $v_{i}$ = initial velocity

$v_{f}$ = final velocity

h = height

g = acceleration due to gravity

$F_{d}$=drag force

m = mass of the car

d = distance

Put the value into the formula

$\dfrac{1}{2}\times8.10\times10^{3}\times25.8=\dfrac{1}{2}\times8.10\times10^{3}\times13.1+8.10\times10^{3}\times9.8\times12.5+F\times3.90\times10^{2}$

$F=\dfrac{\dfrac{1}{2}\times8.10\times10^{3}\times25.8-\dfrac{1}{2}\times8.10\times10^{3}\times13.1-8.10\times10^{3}\times9.8\times12.5}{3.90\times10^{2}}$

$F=-2412.34\ N$

$|F|=2412.34\ N$

Hence, The magnitude of the average drag force is 2412.34 N.