How Do You Calculate Forces on a Fairground Ride Ramp?

[Chris08]

Homework Statement

Hi i have a question I've been doing and need to know if i am completely wrong (or not) any help woul be really appreciated.

A fairground ride ends with the car moving up aramp at a slope of 20degrees to the horizontal

Homework Equations

a)The car carrying its maximum load of passengers has a total weight of 6.8kN.Show that the component of the weight acting parallel to the ramp is about 2.3kN.

b) The mass of the fully loaded car is 690kg. Show that the force in part a will deccelerate the car at about 3.3ms^2

c) The car enters the ramp at 22ms^1.Calculate the minimum length that the ramp must be in order for the car to stop before it reaches the end.Neglect the length of the car.

The Attempt at a Solution

Ok for a) i did tan20degrees =opp/adj so tan20degreesx6.8kn=adj =2.47kN (horizontal component).

b) i did parallel component = f=ma so a=2.3/690 = 3.3ms^2

c) i used v=u^2+2as = 0= (22ms^1)^2 +2(-3.3ms^2x s)
=484=-6.6xs so s =484/6.6 = 73m.

d) i am not to sure about this one can anybody help me with this please?

Thanks,Chris

 

[qspeechc]

Part a) asks for the parallel component
c) would be solved easier with conservation of energy.
You did not post question d).