A 515-kg roller coaster is at the bottom of a loop with a radius of 10 m. If the speed at the Bottom of the loop is 20 m/s, what is the force of the track pushing up on the vehicle at this
point?

20600N

Explanation:

Given parameters:

Mass of the roller coaster  = 515kg

radius  = 10m

Speed  = 20m/s

Unknown:

Force of the track pushing up on the vehicle  = ?

Solution:

The force pulling the roller coaster along the track is the centripetal force.

To find the centripetal force, use the expression below;

  F  =

where;

  m is the mass

  v is the speed

  r is the radius

Now insert the parameters and solve;

              F =  = 20600N

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west