Question

Everything being equal, including speed and mass, like a man walking up a slope, does the amount of energy expended on say an $10\%$ inclined slope double with an $20\%$inclined slope? If travelling at the same speed and everything else the same?
In other words, is energy expended related arithmetically and constant according to the slope of the incline? And resistance, or is it proportional to the slope by some other factor, geometric increase?

Answer 1

Understand the problem and create a diagram based on the context of the solution. Find out the rate of change of energy with the change of inclination angle.

Complete step by step answer:

Write the expression for the potential energy $E$
$E = mgy$
Here, $y$ is his height above the bottom of the hill and $g$ is the acceleration due to gravity.
Write the expression for the rate at which his energy is changing
$\dfrac{{dE}}{{dt}} = mg\dfrac{{dy}}{{dt}}{\text{ }}$
Here, $P$ is the power
Substitute $V\sin \theta$ for $\dfrac{{dy}}{{dt}}$

$P = \dfrac{{dE}}{{dt}}$

$\Rightarrow mgV\sin \theta {\text{ }}$ …… (1)

Here, $m$ is the mass of the person and $\theta$ is the angle of inclination.
Understand that, the velocity, mass of the person remains same for the both cases of inclination provided.
Consider the inclination of $10\%$
Write the expression for the angle of inclination
${\theta _1} = {\tan ^{ - 1}}(m)$

Here, $m$ is the slope of the inclined plane and ${\theta _1}$ is the angle of inclination.

Substitute $10\%$ for $m$

${\theta _1} = {\tan ^{ - 1}}(10\% )$

$\Rightarrow {\tan ^{ - 1}}(\dfrac{{10}}{{100}})$

$\Rightarrow 5.71$

Consider the inclination of $10\%$

Write the expression for the angle of inclination

${\theta _2} = {\tan ^{ - 1}}(m)$

Here, $m$ is the slope of the inclined plane and ${\theta _2}$ is the angle of inclination.

Substitute $10\%$ for $m$

${\theta _2} = {\tan ^{ - 1}}(20\% )$

$\Rightarrow {\tan ^{ - 1}}(\dfrac{{20}}{{100}})$

$\Rightarrow 11.31$

Rewrite the equation (1) for ${\theta _1} = 5.71$

${P_1} = \dfrac{{dE}}{{dt}}$

$\Rightarrow mgV\sin {\theta _1}{\text{ }}$

Rewrite the equation (1) for ${\theta _2} = 11.31$

${P_2} = \dfrac{{dE}}{{dt}}$

$\Rightarrow mgV\sin {\theta _2}{\text{ }}$

Divide ${P_1}$ by ${P_2}$

$\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{\text{sin}}{\theta _1}}}{{\sin {\theta _2}}}$

Substitute $5.71$ for ${\theta _1}$ and $11.31$ for ${\theta _2}$

$\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{\text{sin(}}5.71)}}{{\sin (11.31)}}$

$\therefore 0.57$

Therefore, the power or the rate of change of energy with respect to time for $10\%$ is almost half of the rate of change of energy with respect to time for $20\%$.

Note: Diagram is created using the context of the problem, and the vertical component of velocity is considered for the vertical displacement by the person climbing up the slope. Expression for the angle of inclination is used to determine the angle of inclination and change in the rate of energy is calculated using the expression for the potential energy differentiated with respect to time.