Question

A man weighing $70\mathrm{kg}$ carries a weight of $10\mathrm{kg}$ to the top of the tower $100\mathrm{m}$ high. Calculate the work done.

Answer 1

Step 1: Given

Mass of man = $70\mathrm{kg}$ and mass of weight carried = $10\mathrm{kg}$

Thus, total mass = $70+10=80\mathrm{kg}$

Height = $100\mathrm{m}$

Step 2: Calculate the work done

Work done is equal to the stored potential energy. Thus,

$\mathrm{W}=\mathrm{PE}$ …… (Here, $\mathrm{W}$ is work done, $\mathrm{PE}$ is the potential energy)

$\mathrm{W}=\mathrm{mgh}$ …… $\mathrm{PE}=\mathrm{mgh}$…… (Here, $\mathrm{m}$ is the total mass, $\mathrm{g}$ is the acceleration due to gravity and $\mathrm{h}$ is the height)

Substitute $80\mathrm{kg}$ for the mass, $9.8\mathrm{m}/{\mathrm{s}}^2$ for the acceleration due to gravity and $100\mathrm{m}$ for the height. Thus,

$\mathrm{W}=80\times 9.8\times 100$

$\mathrm{W}=78400\mathrm{J}$

Thus, the work done by a man weighing $70\mathrm{kg}$ carrying a weight of $10\mathrm{kg}$ to the top of the tower $100\mathrm{m}$ high is $78400\mathrm{J}$ i.e. $78400$ joule.