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Q.

A block m slides down a curved frictionless track, starting from rest. The curve obeys the equation $y = \frac{x^{2}}{2} .$ The tangential acceleration of the block is

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a

$\frac{gx}{\sqrt{x^{2} + 1}}$

b

g

c

$\frac{gx}{\sqrt{x^{2} + 4}}$

d

$\frac{g}{2}$

answer is D.

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Detailed Solution

$\frac{dy}{dx} = \tan θ = x \Rightarrow \sin θ = \frac{x}{\sqrt{x^{2} + 1}}$

Tangential acceleration $= gsin θ = \frac{gx}{\sqrt{x^{2} + 1}}$

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A block m slides down a curved frictionless track, starting from rest. The curve obeys the equation y=x22. The tangential acceleration of the block is