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Q7.7-41 PE

Expert-verifiedFound in: Page 262

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 \(4.93 \times {10^4}{\rm{ N}}\).

**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.

(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 \(223{\rm{ s}}\).

(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 \(24.77{\rm{ MW}}\).

(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 \(4.93 \times {10^4}{\rm{ N}}\).

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