Physics energy problem.?

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M~A~T~H

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A roller coaster travelling without friction is at point A 15m above the ground travelling at 3.00m/s. At point B it is 7m above the ground, what is the speed at point B?
 
Use the law of conservation of energy, i.e.

(KE + PE) @ Point A = (KE + PE) @ Point B

(1/2)m(3)^2 + mg(15) = (1/2)(m)(Vb)^2 + mg(7)

where

m = mass of the roller coaster
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
Vb = velocity of coaster at point B

Since "m" appears on both sides of the equation, it will cancel out and the above simplifies to

(1/2)(9) + (9.8)(15) = (1/2)(Vb)^2 + (9.8)(7)

and solving for Vb,

Vb = 12.88 m/sec

Hope this helps.
 
conservation of energy says
PE+KE=PE+KE
so
mgh1+.5mv^2=mgh2+.5mv^2
since the mass is the same for all it can be eliminated
so
gh1+.5v^2=gh2+.5v^2
the left side of the equation is point A and the right side is point B. so just substitute in to find the values.
 
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