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Q7.7-41 PE

Expert-verified
Found in: Page 262

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) How long would it take a $$1.50 \times {10^5} - {\rm{kg}}$$ airplane with engines that produce $$100{\rm{ MW}}$$ of power to reach a speed of $$250{\rm{ m}}/{\rm{s}}$$ and an altitude of $$12.0{\rm{ km}}$$ if air resistance were negligible? (b) If it actually takes $$900{\rm{ s}}$$, what is the power? (c) Given this power, what is the average force of air resistance if the airplane takes $$1200{\rm{ s}}$$? (Hint: You must find the distance the plane travels in $$1200{\rm{ s}}$$assuming constant acceleration.)

(a) Time required by the aircraft is $$223{\rm{ s}}$$.

(b) Power of the engine is $$24.77{\rm{ MW}}$$.

(c) The average force of resistance is $$4.93 \times {10^4}{\rm{ N}}$$.

See the step by step solution

## Step 1: Definition of Concepts

Power: Power is a scalar quantity which is defined how fast the energy is being used.

Mathematically,

$$P = \frac{W}{T}$$

Here, $$P$$ is the power, $$W$$ is the work done, and $$T$$ is the time.

## Step 2: Find the time required by the air craft

(a)

The work done by the airplane is,

\begin{aligned}W = \Delta {\rm{KE}} + \Delta {\rm{PE}}\\ = \frac{1}{2}m{v^2} + mgh\end{aligned}

Here, $$m$$ is the mass of the airplane $$\left( {m = 1.50 \times {{10}^5}{\rm{ kg}}} \right)$$, $$v$$ is the velocity of the airplane $$\left( {v = 250{\rm{ m}}/{\rm{s}}} \right)$$, $$g$$ is the acceleration due to gravity $$\left( {g = 9.81{\rm{ m}}/{{\rm{s}}^2}} \right)$$, and $$h$$ is the height $$\left( {h = 12.0{\rm{ km}}} \right)$$.

Putting all known values,

\begin{aligned}W &= \left( \begin{aligned}{l}\frac{1}{2} \times \left( {1.50 \times {{10}^5}{\rm{ kg}}} \right) \times {\left( {250{\rm{ m}}/{\rm{s}}} \right)^2} + \\\left( {1.50 \times {{10}^5}{\rm{ kg}}} \right) \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {12.0{\rm{ km}}} \right)\end{aligned} \right)\\ &= \left( \begin{aligned}{l}\frac{1}{2} \times \left( {1.50 \times {{10}^5}{\rm{ kg}}} \right) \times {\left( {250{\rm{ m}}/{\rm{s}}} \right)^2} + \\\left( {1.50 \times {{10}^5}{\rm{ kg}}} \right) \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {12.0{\rm{ km}}} \right) \times \left( {\frac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right)\end{aligned} \right)\\ &= 2.23 \times {10^{10}}{\rm{ J}}\end{aligned}

The time required by the airplane,

$$T = \frac{W}{P}$$

Here, $$P$$ is the power generated by the engine of the airplane $$\left( {P = 100{\rm{ MW}}} \right)$$.

Putting all known values,

\begin{aligned}T &= \frac{{2.23 \times {{10}^{10}}{\rm{ J}}}}{{100{\rm{ MW}}}}\\ &= \frac{{2.23 \times {{10}^{10}}{\rm{ J}}}}{{\left( {100{\rm{ MW}}} \right) \times \left( {\frac{{{{10}^6}{\rm{ W}}}}{{1{\rm{ MW}}}}} \right)}}\\ &= 2.23{\rm{ s}}\end{aligned}

Therefore, the time required by the aircraft is $$223{\rm{ s}}$$.

## Step 3: Power of the airplane

(b)

The power of the airplane is,

$$P = \frac{W}{T}$$

Here, $$W$$ is the work done $$\left( {W = 2.23 \times {{10}^{10}}{\rm{ J}}} \right)$$ and $$T$$ is the time taken $$\left( {T = 900{\rm{ s}}} \right)$$.

Putting all known values,

\begin{aligned}P &= \frac{{2.23 \times {{10}^{10}}{\rm{ J}}}}{{900{\rm{ s}}}}\\ &= 24.77 \times {10^6}\\ &= 24.77{\rm{ MW}}\end{aligned}

Therefore, the power of the engine is $$24.77{\rm{ MW}}$$.

## Step 4: Find the average frictional force

(c)

The work done by the airplane assuming friction,

$$W' = PT'$$

Here, $$P$$ is the power of airplane assuming air resistance or friction $$\left( {P = 24.77{\rm{ MW}}} \right)$$, and $$T'$$ is the time $$\left( {T' = 1200{\rm{ s}}} \right)$$.

Putting all known values,

\begin{aligned}W' &= \left( {24.77{\rm{ MW}}} \right) \times \left( {1200{\rm{ s}}} \right)\\ &= \left( {24.77{\rm{ MW}}} \right) \times \left( {\frac{{{{10}^6}{\rm{ W}}}}{{1{\rm{ MW}}}}} \right) \times \left( {1200{\rm{ s}}} \right)\\ &= 2.97 \times {10^{10}}{\rm{J}}\end{aligned}

The net work done is given as,

$$W' = W + {W_f}$$

Here, $$W$$ is the work done $$\left( {W = 2.23 \times {{10}^{10}}{\rm{ J}}} \right)$$, and $${W_f}$$ is the work done by the friction.

The work done by the friction is,

$${W_f} = W' - W$$

Putting all known values,

\begin{aligned}{W_f} &= \left( {2.97 \times {{10}^{10}}{\rm{ J}}} \right) - \left( {2.23 \times {{10}^{10}}{\rm{ J}}} \right)\\ &= 7.4 \times {10^9}{\rm{ J}}\end{aligned}

The first equation of the motion is,

$$v = u + aT$$

Here, $$v$$ is the final velocity $$\left( {v = 250{\rm{ m}}/{\rm{s}}} \right)$$, $$u$$ is the initial velocity $$\left( {u = 0} \right)$$, a is the acceleration, and $$T$$ is the time $$\left( {T = 1200{\rm{ s}}} \right)$$.

From equation (1.1),

$$aT = v - u$$

The second equation of motion the distance traveled by the airplane is,

$$s = uT + \frac{1}{2}a{T^2}$$

Using equation (1.2),

$$s = uT + \frac{1}{2}\left( {v - u} \right)T$$

Putting all known values,

\begin{aligned}s &= \left( 0 \right) \times \left( {1200{\rm{ s}}} \right) + \frac{1}{2}\left( {\left( {250{\rm{ m}}/{\rm{s}}} \right) - \left( 0 \right)} \right) \times \left( {1200{\rm{ s}}} \right)\\ &= 1.5 \times {10^5}{\rm{ m}}\end{aligned}

The work done by friction is,

$${W_f} = Fs$$

Here, F is the average frictional force.

The average frictional force is,

$$F = \frac{{{W_f}}}{s}$$

Putting all known values,

\begin{aligned}F &= \frac{{7.4 \times {{10}^9}{\rm{ J}}}}{{1.5 \times {{10}^5}{\rm{ m}}}}\\ &= 4.93 \times {10^4}{\rm{ N}}\end{aligned}

Therefore, the required average force of resistance is $$4.93 \times {10^4}{\rm{ N}}$$.