Question

A soap-box derby race car sits on the starting ramp. The ramp is elevated at an angle of 8.5 degrees. The angle between the gravity vector and the ramp surface is a . The force of gravity on the car is 1225 N. The normal force of the track on the car is 1210 N. The frictional force of the track on the car is 15.2 N. What is the net force on the car?

Answer 1

`F_(net) =165.87\ N`

Step-by-step explanation:

given,

angle of inclination with the horizontal = 8.5°

force of gravity = 1225 N

normal force on car = 1210 N

frictional force = 15.2 N

net force = ?

equating horizontal forces

`F_x = f - mg sin \theta`

`F_x = 15.2 - 1225 sin 8.5^0`

F_x = -165.87 N

equating vertical forces

`F_y = N - mg cos\theta`

`F_y = 1210 - 1225 cos 8.5^0`

F_y = -1.54 N

net force

`F_(net) = √(F_x^2+F_y^2)`

`F_(net) = √(-165.87^2+-1.54^2)`

`F_(net) = √(27515)`

`F_(net) =165.87\ N`