Question

a body of mass 20 kg is dropped from a height of 100 M find its kinetic energy and potential energy 1 first second 2 second second 3 third second

Answer 1

hey your answer here

A body of mass 20kg is dropped from a height of 100m. ... Therefore the kinetic energy for the first second will be, ... KE1=12×20 kg×(9.8 m/s2g)2(1 s)2KE1= 960.4 ...

I hope it helps

Answer 2

Answer:

The kinetic energy at the first, second, and third seconds are 1000J,4000J, and 9000J respectively. And the potential energy at the first, second, and third seconds are 19000J,16000J, and 11000J respectively.

Explanation:

Here we will take acceleration as acceleration due to gravity i.e.

a=g=10ms⁻²

and the initial velocity of the body will be zero.

At t=1 second

$S_1=\frac{1}{2}gt_1^2=\frac{1}{2}\times 10\times 1^2=5m$

$v_1=gt_1=10\times 1=10ms^-^1$

Kinetic Energy,

$K_1=\frac{1}{2}mv_1^2=\frac{1}{2}\times 20\times 10^2=1000J$

Potential energy,

$U_1=mg(h-S_1)=20\times 10\times(100-5)=19000J$

At t=2 second

$S_2=\frac{1}{2}gt_2^2=\frac{1}{2}\times 10\times 2^2=20m$

$v_2=gt_2=10\times 2=20ms^-^1$

Kinetic Energy,

$K_2=\frac{1}{2}mv_2^2=\frac{1}{2}\times 20\times 20^2=4000J$

Potential energy,

$U_2=mg(h-S_2)=20\times 10\times(100-20)=16000J$

At t=3 second

$S_3=\frac{1}{2}gt_3^2=\frac{1}{2}\times 10\times 3^2=45m$

$v_3=gt_3=10\times 3=30ms^-^1$

Kinetic Energy,

$K_3=\frac{1}{2}mv_3^2=\frac{1}{2}\times 20\times 30^2=9000J$

Potential energy,

$U_3=mg(h-S_3)=20\times 10\times(100-45)=11000J$

Hence, the kinetic energy at the first, second, and third seconds are 1000J,4000J, and 9000J respectively. And the potential energy at the first, second, and third seconds are 19000J,16000J, and 11000J respectively.