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Question: A man of 60 kg carries a stone of 20 kg to a height of 30m. Calculate the work done by the man....

A man of 60 kg carries a stone of 20 kg to a height of 30m. Calculate the work done by the man.

Explanation

Solution

This type of question is easy to answer as they are directly derived from the formula. We just need to apply them and substitute the value. Work done is asked and a man is carrying a stone to a height of 30m which gives clear vision of using potential energy which is W=mghW = mgh where, m is the mass, g is the gravity, and h is the height.

Complete step by step answer:
Step 1:
Before proceeding to the question let us see some important definition for the solution purpose:
Definition of work done: Work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement. If the force is being exerted at an angleθ\theta to the displacement, the work done is W=fdcosθfd\cos \theta
To show that work done (kinetic energy) is equal to the increase in potential energy, turn the scenario around: Work gained (increase in KE) is equal to decrease in PE, as a weight of mass m drops a height of h.
If an object's kinetic energy doesn't change, then no work is done. Another Equation for Calculating Work: Work = Mass × Gravity × Height or MGH and is measured in Joules.
Step 2:
We are given that: mass of a man is mm=60Kg{m_m} = 60Kg
Mass of the stone is ms=20Kg{m_s} = 20Kg
Height equal to 30m
We need to calculate the work done by the man.
Now the work done is W=mghW = mgh where, m is the mass, g is the gravity, and h is the height.
Here m =mm+ms{m_m} + {m_s}
Substituting the value in work done we get W=(mm+ms)×g×h\left( {{m_m} + {m_s}} \right) \times g \times h
W=(60+20)10×30\Rightarrow W = \left( {60 + 20} \right)10 \times 30 , which on solving gives work done equal to 24000 J
Hence the work done is 24000 J

Note: The formula of work done is expressed in joules and can find with the help of kinetic or potential energy. Here we have used potential energy because of two reasons. First is there is no velocity mentioned which is necessary in kinetic and also here mention that a man is lifting a stone which gives us a clear option to choose potential energy.