Question

Show that when a body is dropped from a certain height,the sum of its kinetic energy at any instant during its fall is constant

Answer 1

Answer:

Sum of kinetic energy and potential energy will remain constant for freely falling object

Explanation:

If a body is dropped from Height "H"

Now the initial gravitational potential energy of the body is given as

$U = mgH$

initial kinetic energy = 0

now let say it is dropped by distance "y"

so its height from the ground is "H - y"

so new potential energy is

$U = mg(H - y)$

by kinematics we can find its speed as

$v^2 - 0 = 2gy$

now its kinetic energy is given as

$K = \frac{1}{2}m(2gy) = mgy$

now final sum of kinetic energy and potential energy is

$U + K = mgy + mg(H - y)$

$U + K = mgH$

so here sum of energy is same as that of initial energy

#Learn

Topic : Mechanical energy conservation law

https://brainly.in/question/7771051

Answer 2

Answer:

Hope this may help you..