Question

In case of freely falling body, the ratio ofkinetic energy at the end of the third second to increase
kinetic energy in the next three second is​

Answer 1

Answer:

Kinetic Energy : 1/2mv².

Explanation:

The ratio of the kinetic energy at the end of the third second to increase kinetic energy in the next 3 second is 1 is to 3.

The freely falling body has 1 by 9 times in the of the kinetic energy (1 : 9).

Then the next kinetic energy in the next three seconds is (3 : 1)

soso the ratio of kinetic energy at the end of the third second to increase in kinetic energy next 3 second is

$= 3 \times \frac{1}{9}$

$= \frac{1}{3}$

Ans: 1:3

Answer 2

The ratio will be 1 : 3

we can find the velocity of the body after the first three seconds through 1st law of motion as

V= u+ at

here u will be zero as the body is free falling

v= g(3)= 3g

v = 3g m/s

Kinetic energy will be K1 = 1/2 mv^2

K1 = 1/2m ( 3g) ^2 = 9/2(mg^2)

Velocity after the next 3 seconds will be

v = g(6)

v= 6g m/s

kinetic energy will be 1/2mv^2

1/2 m (6g)^2

Change in the kinetic energy of the body in between the 3rd and 6th second will be

K2 = 1/2m ( (6g)^2 - (3g) ^2)

K2 = 1/2m( 27g^2)

K2 = 27/2 mg^2

the required ratio will be K1/K2

(9/2mg^2)/(27/2mg^2) = 9/27

the ratio of kinetic energy at the end of the third second to increase the kinetic energy in the next three seconds is 1/3.