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

The magnitude of the average drag force is 2412.34 N.

Step-by-step explanation:

Given that,

Mass of car
`m=8.10*10^(-3)\ kg`

Velocity v = 25.8 m/s

Distance
`d= 3.90*10^(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)`

`(1)/(2)mv_(i)^2=\frac{}{}mv_(f)^2+mgh+F* 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

`(1)/(2)*8.10*10^(3)*25.8=(1)/(2)*8.10*10^(3)*13.1+8.10*10^(3)*9.8*12.5+F*3.90*10^(2)`

`F=((1)/(2)*8.10*10^(3)*25.8-(1)/(2)*8.10*10^(3)*13.1-8.10*10^(3)*9.8*12.5)/(3.90*10^(2))`

`F=-2412.34\ N`

`|F|=2412.34\ N`

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