Question

An object of mass 500 g falls from a height of 2m. If $g=9.8m/s^2$, what is its kinetic energy just before touching the ground?

• A. Zero
• B. 19.6 J
• C. 9.8 J
• D. 9800 J

Answer 1

The correct option is C 9.8 J

We know,

Acceleration, $a=\frac{v^2-u^2}{2s}$

where $v$ and $u$ are final and initial velocities respectively & $s$ is the displacement.

Here, $g=\frac{v^2-u^2}{2h}.$ Also, $u=0$

$⟹$ Velocity when it touches ground, $v=\sqrt{2gh}$

$v=\sqrt{2gh}=\sqrt{2\times 9.8\times 2}$

$K.E=\frac{1}{2}m{v}^2=\frac{1}{2}\times \frac{500}{1000}\times (\sqrt{2\times 9.8\times 2}{)}^2$ $=\frac{1}{2}\times \frac{500}{1000}$$\times 2\times 9.8\times 2=9.8J$