Question

A 1000 kg sports car accelerates from 0 m/s to 30 m/s in 11.
What is the average power of the engine?

Answer 1

Concepts and reason

The concepts used in this problem are Newton’s equations of motion, Newton’s second law of motion, work done and the power.

First, calculate the displacement of the car using the Newton’s equation of motion. Then calculate the average power of the engine using the concept of work done and the power.

Fundamentals

Newton’s equations of motion:

The Newton’s equations of motion give the relation between initial speed, final speed, acceleration, distance and time of a motion. These equations describe the motion of an object.

Three equations of motion are:

$\begin{array}{l}\\v - u = at\\\\{v^2} - {u^2} = 2as\\\\s = ut + \frac{1}{2}a{t^2}\\\end{array}$

Here, $v$is the final speed, $u$is the initial speed, $a$is the acceleration, $s$is the distance and $t$is the time.

Work done:

The work done in lifting an object from the ground is the product of force and displacement. The expression of work is:

$W = Fs$

Here, $F$is the force and$s$is the displacement.

Power:

The energy per unit time is the power. The expression for power is:

$P = \frac{E}{t}$

Here, $E$is the energy and $t$is the time.

Newton’s Second Law of motion:

According to Newton’s second law of motion, the force on an object is equal to the product of mass and acceleration. It depends on mass and acceleration.

The expression of net force is:

${F_{{\rm{net}}}} = ma$

Here, $m$is the mass and $a$is the acceleration.

The equations of motion used here are:

$s = ut + \frac{1}{2}a{t^2}$ …… (1)

$v - u = at$ …… (2)

The acceleration of the car is:

$a = \frac{{v - u}}{t}$

Substitute$0{\rm{ m/s}}$for$u$, $30{\rm{ m/s}}$for$v$, $11{\rm{ s}}$for$t$ in equation (3).

$\begin{array}{c}\\a = \frac{{v - u}}{t}\\\\ = \frac{{30{\rm{ m/s}} - 0{\rm{ m/s}}}}{{11{\rm{ s}}}}\\\\ = 2.73{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

The displacement of the car is:

$s = ut + \frac{1}{2}a{t^2}$

Substitute$0{\rm{ m/s}}$for$u$, $2.73{\rm{ m/}}{{\rm{s}}^2}$for$a$, $11{\rm{ s}}$for$t$ in equation (3).

$\begin{array}{c}\\s = \left( {0{\rm{ m/s}}} \right)\left( {11{\rm{ s}}} \right) + \frac{1}{2}\left( {2.73{\rm{ m/}}{{\rm{s}}^2}} \right){\left( {11{\rm{ s}}} \right)^2}\\\\ = 165.2{\rm{ m}}\\\end{array}$

The work done on the engine of the car is equal to the energy of the car according to work-energy theorem.

$E = W$

$E = Fs$ …… (3)

According to Newton’s second law of motion, the force is:

$F = ma$ …… (4)

Substitute equation (4) in equation (3).

$E = mas$

Substitute${\rm{1000 kg}}$for$m$, ${\rm{2}}{\rm{.73 m/}}{{\rm{s}}^{\rm{2}}}$for$g$and$165.2{\rm{ m}}$for$s$.

$\begin{array}{c}\\E = \left( {1000{\rm{ kg}}} \right)\left( {{\rm{2}}{\rm{.73 m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {165.2{\rm{ m}}} \right)\\\\ = 450996\;{\rm{J}}\\\end{array}$

The power is:

$P = \frac{E}{t}$

Substitute$450996\;{\rm{J}}$for$E$and$11\;{\rm{s}}$for$t$.

$\begin{array}{c}\\P = \frac{{450996\;{\rm{J}}}}{{11\;{\rm{s}}}}\\\\ = 40999.6{\rm{ W}}\\\\ = {\bf{4}}{\bf{.1 \times 1}}{{\bf{0}}^{\bf{4}}}{\bf{ W}}\\\end{array}$

Ans:

The average power of the engine is${\bf{4}}{\bf{.1 \times 1}}{{\bf{0}}^{\bf{4}}}{\bf{ W}}$.