Question

A boy weighing $50\,kg$ eats bananas. The energy content of banana is $1000\,cal$, if this energy is used to lift the boy from ground then the height through which he is lifted is:
A. $8.57\,m$
B. $17\,m$
C. $6.57\,m$
D. $5.57\,m$

Answer 1

Hint: The energy required to lift the boy to a height h will be equal to the value of potential energy at that height. The energy content of the banana is given. By equating this with the equation for potential energy we can find the height to which the boy can be lifted using this energy.

Complete step by step answer:
It is given that a boy weighing $50\,kg$ eats bananas.
So, mass of the boy, $m = 50\,kg$
The energy content of the banana is given as $1000\,cal$
We know that one calorie is $4.2\,J$ .
Hence the given energy content in joules is
$ \Rightarrow E = 1000 \times 4.2\,J$
$ \Rightarrow E = 4200\,J$
We need to find the height to which the man can be lifted using this energy.
Let us suppose that the height through which he is lifted is h.
We know that the energy at a height is the gravitational potential energy given as
$PE = mgh$
Where, m is the mass, g is acceleration due to gravity and h is the height.
Let us equate the energy content of bananas to the value of this potential energy.
$ \Rightarrow E = mgh$
From this we can find the value of h as
$ \Rightarrow h = \dfrac{E}{{mg}}$
We know that the value of g is $9.8\,m/{s^2}$
Let us substitute all the given values in this equation.
$ \Rightarrow h = \dfrac{{4200}}{{50 \times 9.8}}$
$ \Rightarrow h = 8.57\,m$
This is the height to which the boy can be lifted using the given energy content of the banana.
So, the correct answer is option A.

Note: Remember that in order to move an object to vertical height we should do a work against the gravitational force of attraction of the earth. This work is stored as a gravitational potential energy at that height given as $PE = mgh$, where, m is the mass of the object, h is the height and g is the acceleration due to gravity. It is the energy stored by a body by virtue of its vertical position.