Question

A man of $60\,Kg$ gets $1000\,cal$of heat by eating $5$ mangoes. His efficiency is $28\,$%. To what height he can jump by using this energy.
(A) $2\,m$
(B) $20\,m$
(C) $28\,m$
(D) $0.2\,m$

Answer 1

Hint: Use the work done and energy stored to find the work can be done by the man after gaining some energy. The substitute the value of work done in the gravitational potential energy formula to know the value of the height of the jump.

Useful formula:
(1) Work done in terms of efficiency is
$W = E \times \eta $
Where $W$ is the work done, $E$ is the energy stored by the man after eating mangoes and $\eta $ is the efficiency of the man.
(2) Work done by the man is
$W = mgh$
Where $m$ is the mass of the man, $g$ is the gravitational force and $h$ is the height that the man can jump by using the energy.

Complete step by step solution:
Given data from the question are
Weight of the man, $m = 60\,Kg$
Number of mangoes that man ate, $n = 5$
Efficiency of the man, $\eta = 28\,\% $
Energy gained by the man after eating mangoes, $E = 1000\,cal$
By using the formula (1)
$W = E \times \eta $
$W = 1000\,cal \times \left( {\dfrac{{28}}{{100}}} \right)$
Converting calories to joules, such that $1\,cal = 4.184\,J$
$W = 4.184 \times \dfrac{{28}}{{100}} \times 1000$
By doing the arithmetic operation,
$W = 1171.5\,J$
Substituting the value of work done in the formula (2)
$W = mgh$
Substitute the values of mass and gravitational force (9.81) in the above equation
$1171.5 = 60 \times 9.81 \times h$
By further simplification,
$h = 2\,m$
Hence after eating $5$mangoes, the man can jump to a height of $2\,m$.

Thus the option (A) is correct.

Note: The formula (2) $W = mgh$ is used because the work is done against the force of gravity. The man jumps, since jumping is the action that takes place against the earth’s gravitational force. This formula is substituted as Gravitational potential energy $\Delta PE$.