SATHEE: Rotational Motion Question 35

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) $6 .26 \times 1 0^{3}$ Watts at the start

B) $6 .26 \times 1 0^{3}$ Watts at take off

C) $6 .26 \times 1 0^{4}$ Watts at the start

D) $6 .26 \times 1 0^{4}$ 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

$\frac{1}{2} m v^{2} = m g h \Rightarrow v = \sqrt{2} g h$

$h = 1 \Rightarrow v = \sqrt{2} \times 10 \times 1 = 20 m / s$

The maximum power delivered by the muscles

P= $2 \times m g \times v = 2 \times 70 \times 10 \times \sqrt{2} 0$

$P = 6.26 \times 103$ Watts at take off.

Hence the correct answer is $6.26 \times 10^{3}$ Watts at take off.

Rotational Motion Question 35

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Rotational Motion Question 35

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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