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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A roller coaster pictured below starts at point $A$ and speeds down a curved track with speed $u$. Assume that the friction between the roller coaster and the track is negligible even though it always stays in contact with the track. What is the speed of the roller coaster at point $B$?

(A) $ \large\sqrt ( \normalsize u^2 + \large\frac{1}{3} \normalsize gh) \quad$(B) $ \large\sqrt ( \normalsize u^2 + \large\frac{2}{3} \normalsize gh) \quad$ (C) $ \large\sqrt ( \normalsize u^2 + \normalsize gh) \quad$(D) $ \large\sqrt ( \normalsize u^2 - \large\frac{1}{3} \normalsize gh)$

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Total energy at point $A =$ Kinetic + Potential energy $= \large\frac{1}{2}$$mu^2+mgh$
If $u_b$ is the speed at point $B$, total energy at $B$ = $= \large\frac{1}{2}$$mu_b^2+mg (\large\frac{2}{3}$$h)$
From the law of conservation of energy, Total energy at point $A$ = Total energy at point $B$,
$\Rightarrow \large\frac{1}{2}$$mu^2+mgh = \large\frac{1}{2}$$mu_b^2+mg (\large\frac{2}{3}$$h)$
Solving for $u_b \rightarrow u_b = \large\sqrt { \normalsize u^2 + \large\frac{2}{3} \normalsize gh}$
answered Mar 22, 2014 by balaji.thirumalai
edited Jun 13, 2014 by lmohan717

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