
What is the mass of a person if he does 3600 J of work in climbing a tree 6m tall? \[(g = 10m/{s^2})\]
(A) 70 kg
(B) 60 kg
(C) 85 kg
(D) 52 kg
Answer
232.8k+ views
Hint Potential energy is defined as the work done to change the configuration of the body. If a body is raised to a height, some work must be done to change its configuration from the ground state to that elevated state. This work done is stored in the form of potential energy in the body.
Complete step-by-step solution
We are given that a man does 3600J of work in climbing a tree. As there is no other force involved except the force due to gravity, this work done will be stored as potential energy inside the man due to virtue of his height. Potential energy of a body of mass m and height h on planet earth \[(g = 10m/{s^2})\] is given by
\[E = mgh\]
This potential energy is given in the question as 3600J, substituting this in the equation:
\[3600 = m \times 10 \times 6\]
\[m = 60kg\]
Therefore, the mass of the man is 60kg. So, the correct answer is given by option B
Note: Energy is neither created nor destroyed. So, if the man jumps from this height, then all of this potential energy will be converted into kinetic energy. Moments before touching the ground, he will have the highest kinetic energy and his velocity just before touching the ground will be \[v = \sqrt {2gr} \].
Complete step-by-step solution
We are given that a man does 3600J of work in climbing a tree. As there is no other force involved except the force due to gravity, this work done will be stored as potential energy inside the man due to virtue of his height. Potential energy of a body of mass m and height h on planet earth \[(g = 10m/{s^2})\] is given by
\[E = mgh\]
This potential energy is given in the question as 3600J, substituting this in the equation:
\[3600 = m \times 10 \times 6\]
\[m = 60kg\]
Therefore, the mass of the man is 60kg. So, the correct answer is given by option B
Note: Energy is neither created nor destroyed. So, if the man jumps from this height, then all of this potential energy will be converted into kinetic energy. Moments before touching the ground, he will have the highest kinetic energy and his velocity just before touching the ground will be \[v = \sqrt {2gr} \].
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