How Is Kinetic Energy Calculated at Point A in a Loop-the-Loop?

[am08]

[SOLVED] Kinetic Energy

A small block of mass m = 1.5 kg slides, without friction, along the loop-the-loop track shown. the block starts from the point P at rest a distance h = 55.0 m above the bottom of the loop of radius R = 19.0 m. What is the kinetic energy of the mass at the point A on the loop?

So I use conservation of energy K1 + PE1 = K2 + PE2 or .5mv^2(1) + mgR(1) = .5mv^2(2) + mgR(2)

At point A there's 0 KE.

Is this right? Do I just plug numbers in?

 

[Google_Spider]

.5mv^2(1) + mgR(1) = .5mv^2(2) + mgR(2)

This equation seems to be correct, but how did you get KE= zero ??

 

[am08]

assumption...

is it not zero?

 

[Dick]

Noo. At point P there is 0 KE. At point A there is. Tell me what are R(1) and R(2)? Then you can plug numbers in.

 

[am08]

R(1) and R(2) is the radius - 19m

 

[Dick]

Nope. What you are calling mgR should be written mgh where h is the vertical displacement from some fixed position of your choice. h is 55m at P. What is it at A?

 

[am08]

PE = 0 at A

 

[Dick]

PE = 0 at A

You can make that choice. Then what is PE at P?

 

[am08]

Wouldn't it be mgh

 

[Dick]

Let's keep this simple. PE=mgh. They gave you h=55m at P. So PE at P is mg(55m). What's h at A? Then what's PE at A?

 

[am08]

17 m

Got it dick... thanks for sticking with me and helping me out with this problem

 

[Dick]

Nope. 38m. You keep changing origins on me! Pick a zero point and stick with it. If we say h=55m then we are measuring everything from the horizontal line in your picture. You could also measure everything from any other point but that's too confusing. Let's just stick with this one.

 

[Dick]

I gather you got the right answer. Yes, 55m-38m=17m. So mg(17m)=(1/2)mv^2. If you understand it, that's great.