Question

A body of mass 1.5 kg slide down a curved track which is quadrant of a circle of radius 0.75 meter. All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is ..... $(g=10m/s^2)$

• A. 3.87
• B. 2
• C. 1.5
• D. 0.387

Answer 1

The correct option is A 3.87

If the rest position is considered as particle having zero potential energy, then using conservation of energy, we can write at the rest position, the total energy = 0 and the bottom most position, the total energy is $mgL-m{v}^2/2$. Since energy is conserved, we have $mgL-m{v}^2/2=0⟹v=\sqrt{2gL}$. Substituting the values of L and g, we get $v=\sqrt{2\times 10\times 0.75}=3.87m/s$