What is the minimum kinetic energy of the block at B, which is located at the top of the loop, so that the block can pass by this point without falling off from the track?

Question

asked Mar 26, 2021 139k views

1 Answer

PatrykMilewski · Answer 1 · 2021-03-31T04:12:33+0000

Answer:

{K.E = (1)/(2) mgR

Step-by-step explanation:

For loop-the-loop, to stay at the top of the loop the centripetal force must equal the gravitational force

mg = (mv^2)/(R)

solving for velocity we get:

v = √(gR)

Thus, the minimum kinetic energy required is

K.E = (1)/(2) mv^2

$\boxed{K.E = (1)/(2) mgR}$