Physics: Abead slides without friction around a loop the-loop. the bead is released from a height of 24.5 m from the bottom of the loop-the loop which has a radius 9 m. the acceleration of gravity is 9.8 m/s². what is its speed at point a? answer in units of m/s.

im not a genius Explanation:but if i had to guess i am pretty sure yor answer would be c....

Abead slides without friction around a loop the-loop.

emilypk1998

11.01.2022

$v = 11.29\ m/s$

Explanation:

given,

bead is released from the height = 24.5 m from bottom

radius of the loop = 9 m

acceleration due to gravity = 9.8 m/s

A be the top most point of the loop

Difference of elevation from the top

H = 24.5 - 2 x r

H = 24.5 - 2 x 9

H = 24.5 - 18

H = 6.5 m

now using conservation of energy

KE = PE

$\dfrac{1}{2}mv^2 = m g H$

v^2 = 2 g H

$v = \sqrt{2 g H}$

$v = \sqrt{2 \times 9.8 \times 6.5}$

$v = \sqrt{127.4}$

$v = 11.29\ m/s$

speed at point A is equal to $v = 11.29\ m/s$

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