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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.Conservative force & potential energyEasy2 minQB

PhysicsEasysingle choice

A particle is placed at the point $A$ of a frictionless track $ABC$ as shown in figure. It is gently pushed towards right. The speed of the particle when it reaches the point $B$ is: $($ Take $g = 10 m s^{- 2})$ .

Options:

Answer:

Solution:

Applying conservation of mechanical energy between point $A$ & $B$ , we get

$K E_{A} + U_{A} = K E_{B} + U_{B}$

$\Rightarrow 0 + m g (1)= \frac{1}{2} m v^{2} + m g \times 0.5$

$\Rightarrow v = \sqrt{g} = \sqrt{10} m s^{- 1}$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Conservative force & potential energy

⏱ 2mℹ️ Source: QB

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