Chapter 6, Problem 78GP

To determine

The power required from the patient

Answer to Problem 78GP

Solution: $131.5\text{W}$

Explanation of Solution

Given The mass of the patient is $m=75\text{kg}$ .

The slope of the treadmill is ${12}^0$ .

The velocity is,

  $\begin{array}{l}v=3.1\frac{\text{km}}{\text{h}}\\ v=3.1\times \frac{1000}{3600}\frac{\text{m}}{\text{s}}\\ v=0.86\frac{\text{m}}{\text{s}}\end{array}$

Formula used

The power can be obtained by:

  $P=Fv$

Where, F is the force and vis the velocity.

Calculations

The weight of the person is

  $\begin{array}{l}W=mg\\ W=75\times 9.8\text{N}\\ W=735\text{N}\end{array}$

The component of this weight along the slope is

  $\begin{array}{l}{W}_s=735\times \sin {12}^{\circ }\text{N}\\ W_s=152.8\text{N}\end{array}$

This weight is the force here.

Therefore, the power is,

  $\begin{array}{l}P=Fv\\ P=152.8\times 0.86\text{W}\\ P=131.5\text{W}\end{array}$

This power is more than twice that of a lightbulb rating.

Conclusion:

The power can be calculated from the weight calculation of the patient along the slope.