Question

A 710kg car drives at a constant speed of 23m/s . It is subject to a drag force of 500 N. What power is required from the car's engine to drive the car (a) on level ground? (b) up a hill with a slope of 2.0? ?

Answer 1

Concepts and reason

The concept required to solve this problem is power.

Initially use the equation of power due to force to calculate power required.

Finally calculate the net force in the opposite direction to the motion of car and substitute it in the power equation to calculate the power required from the engine.

Fundamentals

The power due to an applied force to keep an object at constant velocity is given as,

$P = Fv$

Here, $F$ is the applied force, and $v$ is the constant velocity at which object moves.

The parallel component of the force of gravity for an inclined plane is,

${F_{{\rm{g}}\parallel }} = mg\sin \theta$

Here, m is the mass of the object, g is the acceleration due to gravity, and $\theta$ is the angle between the horizontal and the incline.

(a)

Use the equation of power.

Substitute $500{\rm{ N}}$ for $F$, and $23{\rm{ m/s}}$ for $v$ in the equation of power $P = Fv$.

$\begin{array}{c}\\P = Fv\\\\ = \left( {500{\rm{ N}}} \right)\left( {23{\rm{ m/s}}} \right)\\\\ = 11500{\rm{ W}}\\\end{array}$

(b)

Use the equation of parallel component of gravity.

Substitute $710{\rm{ kg}}$ for m, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $2^\circ$ for $\theta$ in the equation ${F_{{\rm{g}}\parallel }} = mg\sin \theta$ and calculate the parallel component of the gravity.

$\begin{array}{c}\\{F_{{\rm{g}}\parallel }} = \left( {710{\rm{ kg}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\sin 2^\circ \\\\ = 243{\rm{ N}}\\\end{array}$

The net force opposing the motion is sum of drag and parallel component of the gravity that is,

$F = {F_{{\rm{g}}\parallel }} + {F_{{\rm{drag}}}}$

Substitute $243{\rm{ N}}$ for ${F_{{\rm{g}}\parallel }}$ and $500{\rm{ N}}$ for ${F_{{\rm{drag}}}}$ in the above equation $F = {F_{{\rm{g}}\parallel }} + {F_{{\rm{drag}}}}$.

$\begin{array}{c}\\F = 243{\rm{ N}} + 500{\rm{ N}}\\\\ = 743{\rm{ N}}\\\end{array}$

Use the power equation.

Substitute $743{\rm{ N}}$ for $F$, and $23{\rm{ m/s}}$ for $v$ in the equation of power $P = Fv$.

$\begin{array}{c}\\P = Fv\\\\ = \left( {{\rm{743 N}}} \right)\left( {23{\rm{ m/s}}} \right)\\\\ = 17089{\rm{ W}}\\\end{array}$

Ans: Part a

The power required from the car’s engine to drive car on level ground is $11500{\rm{ W}}$.

Part b

The power required from the car’s engine to drive car up with a slope of $2^\circ$ is ${\rm{17089 W}}$.