Question

A man throws the bricks to the height of $12m$ where they reach with a speed of $12m/s$. If he throws the bricks such that they just reach this height, what percentage of energy will he save?

• A. $19$%
• B. $76$%
• C. $38$%
• D. $57$%

Answer 1

Answer: (C)

$38$%

Energy required in I case $=mgh+\frac{1}{2}m{V}^2$
Energy required in II case $=mgh$
% Energy saved
$\frac{\Delta E}{E}\times 100$

$=\frac{\frac{1}{2}m{V}^2}{mgh+\frac{1}{2}m{V}^2}\times 100$

$=\frac{\frac{1}{2}\times {12}^2}{\frac{1}{2}\times {12}^2+10\times 12}\times 100$

$=\frac{72}{192}\times 100$

$=38$%