A body of mass 1 kg tied to a string is just able to move in a vertical circular path of the radius 1 m . The difference between the kinetic energies at its highest and lowest positions of the body is

Question

Aicharya · Accepted Answer

Kinetic energy is the energy possessed by a body of mass m due to its velocity

v

.

K E = \frac{1}{2} m v^{2}

At the highest point velocity,

v_{A} = \sqrt{r g}

∴ K E = \frac{1}{2} m v_{A}^{2}

At the lowest point velocity,

v_{B} = \sqrt{5 r g}

K E = \frac{1}{2} m v_{B}^{2}

Difference n

K E = \frac{1}{2} m (v_{B}^{2} - v_{A}^{2}) = \frac{1}{2} m (5 r g - r g)

= 2 m r g = 2 \times 1 \times 1 \times 10 = 20 J

A body of mass 1 kg tied to a string is just able to move in a vertical circular path of the radius 1 m .…

Solution