Question

What is the kinetic energy of Sania of mass 50 kg at C, if position C is 20 m below position A?

• A. $5000J$
• B. $10000J$
• C. $15000J$
• D. $20000J$

Answer 1

Answer: (B)

$10000J$

The formula for kinetic energy KE=$\frac{1}{2}m{v}^2$.

Sania covers the distance of $20m$ with a velocity $v$ and the mass is $50kg$. We make use of the third equation of motion to obtain velocity of the body.
$v^2-u^2=2as$
Here, $u=0,a=g=10$$m/s^2$ and $s=20m$.
$v^2-0=2\times 10\times 20,thatis{v}^2=400$

Kinetic energy = $\frac{1}{2}m{v}^2$.

That is, $\frac{1}{2}m{v}^2$$=\frac{50\times 400}{2}=10000J$

Hence, the kinetic energy of Sania at C is $10,000J$