Recent content by d0rk

This is what I got: PE(final)+KE(final)=PE(initial)+KE(initial) 1/2mv^2 + mg(5) = 0 + mg(15) 1/2v^2 + 5g = 0 + 15g 1/2v^2 = 10g v^2 = 20g mg = mv^2/r g = 20g/r r = 20? Is that correct??

so with the velocity at sqrt(30g)... with that when the cart starts to go over the second hump (B) the velocity needs to equal: mg(5)=1/2mv^2 10g=v^2 v=sqrt(10g) I'm assuming that this equals the velocity at the top of hump B... conservation of energy states that...

ok...now after thinking about it...i solved for the velocity at the bottom of the track right before it hits the hump B... mgh=1/2mv^2 mg(15)=1/2mv^2 30g=v^2 v=sqrt(30g) is this the velocity at the bottom of the track for the first part?

Please check if this is right this is what i have so far v^2/R<g K=U 1/2mv^2=mgh v^2=2gh g>2gh/r i put my reference line at the bottom which = 0 the car starts at the top at 15m high then goes down to 0 then back up another hump, point B, at 5.0meters. for h do i subtract...

without any mass how would i be able to incorporate the centripetal force...

if its a circular motion problem shouldn't the net force be 0?..and there is no loop in the diagram...its just point a at 15.0 meters that goes down to 0 meters then back up another hump 5 meters tall...

I know Normal force will be consistent throughout...so should I use the Work Energy Theorem for the external forces...because I know it can't just be mg...

the question is: The figure below shows the plan for a proposed rollercoaster track. Each car will start from rest at point A and will roll with negligible friction. For safety, it is important that there be at least some small positive normal force (a push) exerted by the track on the car...