Question

A ball is dropped from height 10m find the height at which pe=2ke

Answer 1

check the attachment

Answer 2

Answer:

The height at which the Potential Energy is twice  the kinetic Energy is 20/3 m

Explanation:

Given :

Height from which the ball is dropped (H) = 10m

Also, Potential Energy (P.E.) = 2* Kinetic Energy (K.E.)

To Be Calculated :

The height at which P.E. = 2*K.E

Solution:

Let the point at which P.E. = 2*K.E be A,

Then At A, Potential Energy = mgH= mgH

Now, Kinetic Energy :

$K.E. = \frac{1}{2} mv^{2}$

Now, using equations of motion, we have:
$v^{2} =u^{2} +2gh$

Now, for point A, h = 10 - H

and initial velocity (u) = 0 m/s

So,

$v^{2} = 2g(10-H)\\\\v =\sqrt{2g(10 -H)}$

Now, it is given that,

P.E. = 2*K.E

$10mH = 2*\frac{1}{2} *m*2g(10-H)\\\\mgH = 2mg (10-H)\\\\H = 2(10-H)\\\\H = 20 -2H\\\\3H = 20$

Thus, H = 20/3 m

Hence, the height at which the P.E. = 2*K.E is 20/3 m

To read more about Energy, visit

https://brainly.in/question/49270374

https://brainly.in/question/14571194

#SPJ3