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A roller-coaster car has a mass of 1200 kg when fully
loaded with passengers. As the car passes over the top of a circular
hill of radius 18 m, its speed is not changing. At the top of the
hill, what are the (a) magnitude FN and (b) direction (up or
down) of the normal force on the car from the track if the car’s
speed is v 11 m/s? What are (c) FN and (d) the direction if v
14 m/s?

Respuesta :

antonsandiego
I am pretty sure that this solving of a will help you 
a) As far as I am concerned, normal force at the top of the hill  has a formula :[tex] Fc Fn=mg-Fc Fn=mg-mv2/r[/tex] [tex]Fn=1200*9.8-1200(11^2)/18[/tex] [tex]Fn=3693 [/tex]which is up, 'cause it is a positive one.
As you can see forse must be equal to themass (x)gravity - the centripetal force.
the second one (B) -
[tex]m=1200kg r=18m v=11m/s F(c)=(mv^2)/r[/tex]

[tex]=(1200x11^2)/18 .. =(1200x121)/18[/tex][tex]=145200/18 ..=8066.6666[/tex] ..=8100N
And now we can find the solution of C and D[tex]M=1200kg r=18m v= 14m/s[/tex]
[tex]F(c)=(mv^2)/r[/tex]
[tex]......=(1200x14^2)/18 ......=(1200x196)/18[/tex]
[tex]......=(235200)/18 ......=13066.666[/tex]
=13000N 
I do hope you will agree with my thoughts and that you find it helpful.

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