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A $75.0 \mathrm{~kg}$ firefighter climbs a flight of stairs $28.0 \mathrm{~m}$ high. How much work does he do?

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Here given that the mass of the firefighter is $m = 75\,\mathrm{kg}$ .

Height $h = 28\,\mathrm{m}$ .

Here both force $F$ and displacement $d$ are in the same direction, hence the angle $\theta = 0^{\circ}$ .

As we know the work done on an object by a constant force is defined to be the product of the magnitude of the displacement times the component of the force parallel to the displacement and is given by

\begin{align*} W = F_{\left| \right|}\,d\tag{1} \end{align*}

Where $F_{\left| \right|}$ is the component of the constant force $\vec{F}$ parallel to the displacement $\vec{d}$ . We can also write

\begin{align*} W = F\,d\,\cos\theta\tag{2} \end{align*}

where $F$ is the magnitude of the constant force, $d$ is the magnitude of the displacement of the object, and $\theta$ is the angle between the directions of the force $F$ and the displacement $d$ .

So from equation (2), the work done will be

\begin{align*} W & = F\,d\,\cos0^{\circ}\\ & = F\,d\tag{3} \end{align*}

Here the force required to lift the firefighter is

\begin{align*} F = m\,g\tag{4} \end{align*}

So from equations (3) and (4), we have

\begin{align*} W & = m\,g\,d\\ & = 75\,\mathrm{kg} \times\,9.8\,\mathrm{m\,s^{-2}} \times\,28\,\mathrm{m}\\ & = 20580\,\mathrm{kg\,m^2\,s^{-2}} \\ & = 20.58\times 10^{3}\,\mathrm{J} \\ & = 20.58\,\mathrm{kJ} \end{align*}

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