Question

a ball is released from height along the slope and move along a circular track radius R without falling vertically down Show that h=5/2R​

Answer 1

Answer:

Explanation:

=> Velocity of the ball at the lowest point, V = √5gR

=> Potential energy at point, P = Mgh

=> kinetic energy at point, Q = 1 / ⋉ MV²

According to the law of conservation of energy,

potential energy at point P = kinetic energy at point Q

Mgh = 1 / 2 MV²

∴ gh = 1 / 2 (√5gR)²

∴ gh = 1 / 2 (5gR)

h = 5R / 2

Answer 2

=> Velocity of the ball at the lowest point, V = √5gR

=> Potential energy at point, P = Mgh

=> kinetic energy at point, Q = 1 / ⋉ MV²

According to the law of conservation of energy,

potential energy at point P = kinetic energy at point Q

Mgh = 1 / 2 MV²

∴ gh = 1 / 2 (√5gR)²

Read more on Brainly.in - https://brainly.in/question/10799417#readmore