Solved Problem on Work and Energy

A man with a mass of 75 kg climbs a staircase with 15 steps in 10 s. Each step is 20 cm in height and 30 cm in length. Determine:
a) The work done by the weight of a man climbing the ladder;
b) The average power delivered by man.

Problem data:

Mass of the man: m = 75 kg;
Staircase rise time: t = 10 s;
Number of staircase steps: n = 15 steps;
Length of each step: d = 30 cm;
Height of each step: h = 20 cm.

Problem diagram:

Figure 1

Solution

First, we must convert the length and height of the steps given in centimeters (cm) to meters (m) used in the International System of Units (SI).

\[ \begin{gather} d=30\;\cancel{\mathrm{cm}}\times\frac{1\:\mathrm m}{100\;\mathrm{\cancel{cm}}}=0.3\;\mathrm m \\[10pt] h=20\;\mathrm{\cancel{cm}}\times\frac{1\:\mathrm m}{100\;\mathrm{\cancel{cm}}}=0.2\;\mathrm m \end{gather} \]

a) For the calculation of force work weight, it is enough to know the height difference between the initial and final points of man (vertical displacement), for this to only multiply the number of steps by the height of each step

\[ \begin{gather} H=n\;h\\[5pt] H=15\times 0.2\;\mathrm m\\[5pt] H=3\;\mathrm m \tag{I} \end{gather} \]

The weight of man will be

\[ \begin{gather} \bbox[#99CCFF,10px] {W=m g} \end{gather} \]

\[ \begin{gather} W=\left(75\;\mathrm{kg}\right)\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)\\[6pt] W=735\;\mathrm N \tag{II} \end{gather} \]

The work of a force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {\mathscr{W}_{\small F}=F d} \end{gather} \]

where the force will be the weight of man, W, and the distance will be the height he climbed, H

\[ \begin{gather} \mathscr{W}_{\small W}=WH\\[5pt] \mathscr{W}_{\small W}=\left(735\;\mathrm N\right)\times\left(3\;\mathrm m\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathscr{W}_{\small W}=2205\;\mathrm J} \end{gather} \]

Note: The length of the steps does not matter for the calculation of work done by weight, horizontal displacement is perpendicular to weight, and only the force in the direction of the displacement performs work.

b) The average power delivered

\[ \begin{gather} \bbox[#99CCFF,10px] {\mathscr{P}_{avg}=\frac{\mathscr{W}_{\small W}}{\Delta t}} \end{gather} \]

\[ \begin{gather} \mathscr{P}_{avg}=\frac{2205\;\mathrm J}{10\;\mathrm s} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathscr{P}_{avg}=220,5\;\mathrm W} \end{gather} \]

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