A boy weighing \[50\,{\text{kg}}\] climbs up a vertical height of \[100\,{\text{m}}\]. Calculate the amount of work done by him. How much potential energy does he gain (\[g = 9.8\,{\text{m/}}{{\text{s}}^2}\])

Hint: Use the formula for the potential energy of an object. This formula gives the relation between the mass of the object, acceleration due to gravity and height of object from ground. The work done by the boy is equal to the potential energy of the boy at height \[100\,{\text{m}}\]. The gain in potential energy of the boy is the difference of potential energy at height \[100\,{\text{m}}\] and at ground.

Formula used:
The potential energy \[U\] of an object is given by
\[U = mgh\] …… (1)
Here, \[m\] is the mass of the object, \[g\] is acceleration due to gravity and \[h\] is the height of the object from the ground.

Complete step by step solution:
We have given that a body of mass \[50\,{\text{kg}}\] is climbing a vertical height of \[100\,{\text{m}}\].
\[m = 50\,{\text{kg}}\]
\[h = 100\,{\text{m}}\]

While climbing up the vertical height, the work is done by the boy against gravitational force. Hence, the work done by the boy is equal to the potential energy of the boy at height \[100\,{\text{m}}\].
Hence, the work done by the boy is
\[W = mgh\]

Substitute \[50\,{\text{kg}}\] for \[m\], \[100\,{\text{m}}\] for \[h\] and \[9.8\,{\text{m/}}{{\text{s}}^2}\] for \[g\] in the above equation.
\[W = \left( {50\,{\text{kg}}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)\left( {100\,{\text{m}}} \right)\]
\[ \Rightarrow W = 49000\,{\text{J}}\]
Hence, the work done by the boy against gravitational force is \[49000\,{\text{J}}\].

The potential energy gained \[\Delta U\] by the boy is equal to the difference of the potential energy \[{U_f}\] of the boy at height \[100\,{\text{m}}\] and the potential energy \[{U_i}\] of the boy at the ground.
\[\Delta U = {U_f} - {U_i}\]

The potential energy of the boy at ground is zero.
\[{U_i} = 0\,{\text{J}}\]

The potential energy of the boy is equal to the work done in climbing the height \[100\,{\text{m}}\].
\[{U_f} = 49000\,{\text{J}}\]

Substitute \[49000\,{\text{J}}\] for \[{U_f}\] and \[0\,{\text{J}}\] for \[{U_i}\] in the above equation.
\[\Delta U = \left( {49000\,{\text{J}}} \right) - \left( {0\,{\text{J}}} \right)\]
\[ \Rightarrow \Delta U = 49000\,{\text{J}}\]

Hence, the gain in potential energy of the boy is \[49000\,{\text{J}}\].

Note:
One can also solve the same question in another way. The work done by the boy is equal to the negative of change in the potential energy (difference of potential energy of boy at ground and potential energy of boy at height \[100\,{\text{m}}\]) of the boy. The gain in potential energy by the boy is equal to negative of the change in potential energy of the boy.