In a common test for cardiac function (the “stress test”), the patient walks on an inclined treadmill (Fig. 6–46). Estimate the power required from a 75-kg patient when the treadmill is sloping at an angle of 12° and the velocity is \({\bf{3}}{\bf{.1}}\;{{{\bf{km}}} \mathord{\left/{\vphantom {{{\bf{km}}} {\bf{h}}}} \right.} {\bf{h}}}\). (How does this power compare to the power rating of a light bulb?)

FIGURE 6–46 Problem 78.

The power can be calculated by multiplying the applied force with the particle’s velocity. It varies linearly with the applied force.

The mass of the patient is\(m = 75\;{\rm{kg}}\).

The velocity of the treadmill is\(v = 3.1\;{{{\rm{km}}} \mathord{\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.} {\rm{h}}}\).

The angle of inclination is \(\theta = 12^\circ \).

Draw a free-body diagram.

Here, \(F\) is the force applied by the treadmill on the patient, \(N\) is the normal force, \(mg\sin \theta \) is the horizontal component of weight, and \(mg\cos \theta \) is the vertical component of the force.

Apply the equilibrium condition along the vertical direction.

\(\begin{aligned}\sum {F_x} &= 0\\F - mg\sin \theta &= 0\\F &= mg\sin \theta \end{aligned}\)

The power required for the patient can be calculated as shown below:

\(\begin{aligned}P &= Fv\\P &= \left( {mg\sin \theta } \right)v\\P &= \left( {75\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\sin \left( {12^\circ } \right)\left( {\left( {3.1\;{{{\rm{km}}} \mathord{\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.} {\rm{h}}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{km}}}}} \right)\left( {\frac{{{\rm{1}}\;{\rm{h}}}}{{{\rm{3600}}\;{\rm{s}}}}} \right)} \right)\\P &= 131.4\;{\rm{W}}\end{aligned}\)

Thus, the power required for the patient is \(131.4\;{\rm{W}}\).

Compare the power required for the patient withthe power rating of a light bulb.

\(\begin{aligned}\frac{P}{{{P_{{\rm{bulb}}}}}} &= \frac{{\left( {131.4\;{\rm{W}}} \right)}}{{\left( {75\;{\rm{W}}} \right)}}\\P &= 1.7{P_{{\rm{bulb}}}}\end{aligned}\)

Thus, the power required for the patient is 1.7 times that of the power rating of a light bulb.