Question

A body of mass 20 kg is dropped from a height of 6cm.find the kinetic energy of the body just before hitting the ground.(g=10m/s2)​

Answer 1

Answer:

ke=1/2mv^2

Explanation:

m=20kg

v=10m/s^2

ans=1000.

Answer 2

GiveN :-

• A body of mass 20 kg is dropped from a height of 6cm.

To FinD :-

• The kinetic energy of the body just before hitting the ground

SolutioN :-

Here the body is dropped from a height of 6 cm .Firstly change this into SI unit . We can write 6cm as 6/100 m = 0.06 m . The mass of the body is 20 kg . So we know the Kinetic energy as half the product of square of velocity and mass of the body .

$\sf:\implies \pink{ Energy_{(Kinetic)}= \dfrac{1}{2}mv^2}\\\\\sf:\implies Energy_{(Kinetic)} =\dfrac{1}{2}\times 20kg \times (\sqrt{2gh})^2\\\\\sf:\implies Energy_{(Kinetic)} = 10 \times 2\times 10 \times 0.06 m \ J \\\\\sf:\implies\underset{\blue{\sf Required \ Energy }}{\underbrace{\boxed{\pink{\frak{ Energy_{(Kinetic)} = 72 \ Joules }}}}}$