# Please explain this circular movement problem [closed]

I was working over some problems from my physics textbook, and I came across this one, it involves circular movement:

A car with the mass of 1t is moving over a hill with the velocity of 20 m/s. The radius of the hill (which can be viewed as a half-circle) is 100m.

When it comes to the top of the hill, with what force does it act upon the hill?

The correct answer is, apparently, 6kN. How do I solve this?

The situation can be illustrated like this I guess:

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## closed as too localized by David Z♦May 23 '12 at 15:30

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Thanks for the tags! – jcora May 22 '12 at 20:56
What thought have you already made to this? Surely, you have done some! – leftaroundabout May 22 '12 at 21:32
Well, I realize that if the car was just standing there, he would act upon the hill with the force of 10kN. And it's pretty obvious to me that if it was going any faster, the hill would be "pushing him up", and once it would come to the top it would almost fly away. That's why the force is lower when it's moving. – jcora May 23 '12 at 7:43
Hi Bane - as described in our FAQ and homework policy, we don't allow questions that just ask for the solution to a homework-like problem. If you can edit your problem to focus on the specific physical concept that is giving you trouble, I'd be happy to reopen it. (Adding the contents of your last comment to the question would be a great start.) – David Z May 23 '12 at 15:32

Hint: From Newton's 2nd law, $-m g + F_N = -m v^2/r$.