Question

a) Define the term ‘work’. Write the formula for work done on a body when a force acts on the body in the direction of its displacement. Give the meaning of each symbol which occurs in the formula.
b) A person of mass \[50\,{\text{kg}}\] climbs a tower of height \[72\,{\text{meters}}\]. Calculate the work done.

Answer 1

Hint:Recall the definition of the work done in terms of the force and displacement and give the mathematical expression for the wonk done in terms of the force exerted on the object and displacement of an object. Use the relation between the work done and change in potential energy and calculate the work done when the person climbs the tower.
Formula used:
The work done \[W\] is given by
\[W = - \Delta U\] …… (1)
Here, \[\Delta U\] is a change in potential energy of the object.
The potential energy \[U\] of the object is given by
\[U = mgh\] …… (2)
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 answer:
a)
The work is a physical quantity which can be defined as the amount of energy which is transferred to an object or transferred from an object when a force is applied on that object and that force causes the displacement of that object.
One can also define when a force acting on an object causes displacement of that object, work is done.
The formula for work done on a body when a force acts on the body in the direction of its
displacement is as follows:
\[W = Fs\]
Here, \[W\] is the work done, \[F\] is the force acting on the object and \[s\] is the displacement of the object.
b)
We have given that the mass of the person is \[50\,{\text{kg}}\] and the person climbs a tower of
\[72\,{\text{m}}\].
\[m = 50\,{\text{kg}}\]
\[h = 72\,{\text{m}}\]
Initially, the person is on the ground. Hence, the initial potential energy of the person is zero.
\[{U_i} = 0\,{\text{J}}\]
We can calculate the final potential energy \[{U_f}\] of the person using equation (2).
\[{U_f} = mgh\]
Substitute \[50\,{\text{kg}}\] for \[m\], \[9.8\,{\text{m/}}{{\text{s}}^2}\] for \[g\] and
\[72\,{\text{m}}\] for \[h\] in the above equation.
\[{U_f} = \left( {50\,{\text{kg}}} \right)\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)\left(
{72\,{\text{m}}} \right)\]
\[ \Rightarrow {U_f} = 35280\,{\text{J}}\]
Hence, the final potential energy of the person is \[35280\,{\text{J}}\].
According to equation (1), the work done is given by
\[W = - \left( {{U_f} - {U_i}} \right)\]
\[ \Rightarrow W = {U_i} - {U_f}\]
Substitute \[0\,{\text{J}}\] for \[{U_i}\] and \[35280\,{\text{J}}\] for \[{U_f}\] in the above
equation.
\[ \Rightarrow W = \left( {0\,{\text{J}}} \right) - \left( {35280\,{\text{J}}} \right)\]
\[ \Rightarrow W = - 35280\,{\text{J}}\]
Hence, the work done is \[35280\,{\text{J}}\].
The negative sign indicates that the work is done against the gravitational force.

Note: The students should keep in mind that in part a, we have asked to give the formula for work done when the direction of the force acting on an object and displacement of the object is the same. Hence, we have not used the cosine of the angle between the direction force and displacement as the cosine of zero degree is 1. Otherwise, the formula for work done also includes the term cosine of angle between the direction of force and displacement of the object.