Question

A body of mass 2 kg is dropped from height of 10m when it reaches to earth its kinetic energy will be

Answer 1

Given

u=0m/s

v=?

g=9.8m/s^2

h=10m/s

m=2kg

From newton's third law

v^2-u^2=2gh

v^2-0=2×9.8×10

v^2=196

v=14m/s

therefore

kinetic energy = 1/2mv^2

=1/2×2×14×14

=14×14

=196J.

Answer 2

Concept Introduction: Two dimensional motion is a basic part of physics.

Given:

We have been Given:

$m = 2kg \\ u = 0 \\ h = 10m$

To Find:

We have to Find: Kinetic energy.

Solution:

According to the problem, Using the equation

${v}^{2} - {u}^{2} = 2gh \\ {v}^{2} - 0 = 2 \times 10 \times 10 \\ {v}^{2} = 200 \\ v = \sqrt{200} = 14.14m {s}^{ - 1}$

therefore Kinetic energy is

$ke = \frac{1}{2} \times m {v}^{2} \\ = \frac{1}{2} \times 2 \times 199.93 \\ = 199.93 \: joules$

Final Answer: The Kinetic Energy is

$199.93 \: joules$

#SPJ2