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Question: A boy weighing \[50\,{\text{kg}}\] climbs up a vertical height of \[100\,{\text{m}}\]. Calculate the...

A boy weighing 50kg50\,{\text{kg}} climbs up a vertical height of 100m100\,{\text{m}}. Calculate the amount of work done by him. How much potential energy does he gain (g=9.8m/s2g = 9.8\,{\text{m/}}{{\text{s}}^2})

Explanation

Solution

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 100m100\,{\text{m}}. The gain in potential energy of the boy is the difference of potential energy at height 100m100\,{\text{m}} and at ground.

Formula used:
The potential energy UU of an object is given by
U=mghU = mgh …… (1)
Here, mm is the mass of the object, gg is acceleration due to gravity and hh is the height of the object from the ground.

Complete step by step solution:
We have given that a body of mass 50kg50\,{\text{kg}} is climbing a vertical height of 100m100\,{\text{m}}.
m=50kgm = 50\,{\text{kg}}
h=100mh = 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 100m100\,{\text{m}}.
Hence, the work done by the boy is
W=mghW = mgh

Substitute 50kg50\,{\text{kg}} for mm, 100m100\,{\text{m}} for hh and 9.8m/s29.8\,{\text{m/}}{{\text{s}}^2} for gg in the above equation.
W=(50kg)(9.8m/s2)(100m)W = \left( {50\,{\text{kg}}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)\left( {100\,{\text{m}}} \right)
W=49000J\Rightarrow W = 49000\,{\text{J}}
Hence, the work done by the boy against gravitational force is 49000J49000\,{\text{J}}.

The potential energy gained ΔU\Delta U by the boy is equal to the difference of the potential energy Uf{U_f} of the boy at height 100m100\,{\text{m}} and the potential energy Ui{U_i} of the boy at the ground.
ΔU=UfUi\Delta U = {U_f} - {U_i}

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

The potential energy of the boy is equal to the work done in climbing the height 100m100\,{\text{m}}.
Uf=49000J{U_f} = 49000\,{\text{J}}

Substitute 49000J49000\,{\text{J}} for Uf{U_f} and 0J0\,{\text{J}} for Ui{U_i} in the above equation.
ΔU=(49000J)(0J)\Delta U = \left( {49000\,{\text{J}}} \right) - \left( {0\,{\text{J}}} \right)
ΔU=49000J\Rightarrow \Delta U = 49000\,{\text{J}}

Hence, the gain in potential energy of the boy is 49000J49000\,{\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 100m100\,{\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.