Question

A body of 2 kg Falls from rest what will be its kinetic energy during the fall at the end of 2 second assume G barabar 10 M per second square

Answer 1

Mass of body(m)=2kg
g=10m/s
t=2sec
h=10×2=20m
kinetic energy=mgh
=2kg×10m/s×20m
=
$400 kg \: {m}^{2} \: {s}^{ - 1}$

Answer 2

$\Huge{\boxed{\red{\mathbb{QUESTION}}}}$

A body of mass 2 kg falls from rest. What will be it's kinetic energy during the fall at the end of 2 second. Assume acceleration due to gravity 'g' as $10m/s^2.$

$\Huge{\boxed{\pink{\mathcal{ANSWER}}}}$

$\bold{Given\:that,}$

Mass = 2 kg

Initial velocity (u) = 0 m/s

Time = 2 second

Acceleration due to gravity (g) = $10 m/s^2$

But first of all, we've to find the final velocity of the body.

$\bold{By\:using\:1st\:equation\:of\:motion,}$

v = u + at

v = 0 + (10)2

v = 0 + 20

v = 20 m/s

$\bold{By\:using\:the\:K.E.'s\:formula.\:I. E.,}$

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

As here, the Intial velocity is 0.

Therefore, K.E. = $\frac{1}{2} (2 \times 20 \times 20)$

K.E. = $\frac{1}{2} (800)$

K.E. = 400 joules.

$\Huge{\boxed{\red{K.E.\:is\:400\:J}}}$