Question

A power output from a certain experimental car design to be shaped like a cube is proportional to the mass \[m\]of the car. The force of air friction on the car is proportional to\[A{v^2}\], where \[v\]is the speed of the car and A is the cross-sectional area. On a level surface the car has a maximum speed \[{v_{\max }}\]. Assume that all the versions of this design have the same density then \[{v_{\max }}\] is proportional to \[{m^{^{\dfrac{1}{c}}}}\]. Find C.

Answer 1

Hint: In this particular problem, power is proportional to the mass and force of air friction is proportional to\[A{v^2}\]. By using the power formula, we can substitute the values and will find the dependence of power on v. By solving the equations we will try to find the value of C.

Complete step by step answer:
Given, power is proportional to the mass, \[p \propto m\]
and force of air friction is proportional to,\[F \propto A{v^2}\]
As we know, power is the ratio of work done per unit time.
\[P = \dfrac{w}{t}\]
And work done is=\[F \times S\],
After substituting the value of work done-
\[P = \dfrac{{F \times S}}{t}\]
We know, \[\dfrac{s}{t} = velocity\] .
So, \[P = F \times v\]
Now, according to question, \[p \propto m\]
\[F \times v \propto m\]
Now, substitute the value of F that is given in this question.
So,
\[A{v^3} \propto m\] -----------(1)
Now, we need to eliminate the area. So, Consider edge length of the cube is L, So volume of the cube is \[{L^3}\] and mass of the car is given that is- \[m\].If we write mass in terms of density and volume then,
\[m = \rho \times v\]
Here \[m\] is the mass of the car,\[\rho \] is the density and\[v\] is the velocity.
So, we can write-\[m \propto v\]
It means-\[m \propto {L^3}\]
From the above equation, \[L = {m^{\dfrac{1}{3}}}\]
Now put the value of area in equation (1)
We get- \[A{v^3} \propto m\]
\[{L^2}{v^3} \propto m\]
Now put value of L in his equation-
\[
  {m^{\dfrac{2}{3}}}{v^3} \propto m \\
  {v^3} \propto {m^{\dfrac{1}{9}}} \\
 \]
\[{v_{\max }} \propto {m^{\dfrac{1}{9}}}\]
According to question-
\[{v_{\max }} \propto {m^{\dfrac{1}{c}}}\]

Therefore, C=9.

Note: Power generally refers to the rate of doing work. When work is done on an object then energy is transferred. The rate at which this energy is transferred is called power.