A body of mass 1 kg is moving in a vertical circular path of radius 1 m. The difference between the kinetic energies at its highest and lowest position is :

Question

A body of mass 1 kg is moving in a vertical circular path of radius 1 m. The difference between the kinetic energies at its highest and lowest position is : (A) 20 J (B) 10 J (C)  4√5J  (D) 10 (√5-1)J

Tardigrade Admin · 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} = r g

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

At the lowest point velocity,

v_{B} = 5 r g

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

Difference in

K E, Δ K = \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 1 \times 10

= 20 J

Q. A body of mass $1 k g$ is moving in a vertical circular path of radius $1 m$ . The difference between the kinetic energies at its highest and lowest position is :

20 J

10 J

$45 J$

10 $(5 - 1) J$

Q. A body of mass 1kg is moving in a vertical circular path of radius 1m. The difference between the kinetic energies at its highest and lowest position is :

20 J

10 J

45​J

10 (5​−1)J

Q. A body of mass $1 k g$ is moving in a vertical circular path of radius $1 m$ . The difference between the kinetic energies at its highest and lowest position is :

$45 J$

10 $(5 - 1) J$