Rotational Motion Question 35
Question: A 70 Kg man leaps vertically into the air from a crouching position. To take the leap the man pushes the pushes the ground with a constant force F to raise himself. The center of gravity rises by 0.5 m before he leaps. After the leap the c.g. rises by another 1 m. The maximum power delivered by the muscles is: (Take g = 10 $ m{s^{2}} $ ).
[JEE ONLINE 23-04-2013]
Options:
A) $ 6\text{.26}\times 1{0^{3}} $ Watts at the start
B) $ 6\text{.26}\times 1{0^{3}} $ Watts at take off
C) $ 6\text{.26}\times 1{0^{4}} $ Watts at the start
D) $ 6\text{.26}\times 1{0^{4}} $ Watts at take off
Show Answer
Answer:
Correct Answer: B
Solution:
Mass of man m = 70kg
Man pushes the ground with force F to take the leap and his center of gravity rises by 1m after the leap.
Law of conservation of energy is applied .
Kinetic energy initially = potential energy finally
$\frac{1}{2}mv^2=mgh⇒v=\sqrt2gh$
$h=1⇒v=\sqrt2×10×1=20m/s$
The maximum power delivered by the muscles
P=$2×mg×v=2×70×10×\sqrt20$
$P=6.26×103$ Watts at take off.
Hence the correct answer is $6.26×10^3$ Watts at take off.
