A body released from rest slides along a frictionless track $A C$ . Its speed at $C$ is

\sqrt{g h}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{\frac{g h}{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 \sqrt{g h}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{2 g h}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\sqrt{2 g h}$
$∵$ Total mechanical energy is conserved
Initial K.E + Initial P.E = Final K.E + Final P.E
So, the sum of kinetic energy $(K E_{A})$ and potential energy $(P E_{A})$ at point $A$ will be equal to the sum of kinetic energy $(K E_{C})$ and potential energy $(P E_{C})$ at point $C$
$\Rightarrow K E_{A} + P E_{A} = K E_{C} + P E_{C}$
$\Rightarrow 0 + m g h = \frac{1}{2} m v_{C}^{2} + 0$ [ $∵ v_{A} = 0$ So $K E_{A} = 0$ and base of incline is taken as reference]
$\Rightarrow v_{C} = \sqrt{2 g h}$
Velocity at point $C$ is $\sqrt{2 g h}$ .

Suggest Corrections

Similar questions

A C

C

A block of mass m slides along a frictionless surface as shown in the figure. If it released from rest at A, what is its speed at B?

2 k g

1 m

(g = 10 m / s^{2})

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program