Question

An airplane of mass $\mathbf{1}\mathbf{.50}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{}\mathbf{kg}$ is in level flight, initially moving at 60.0 m/s. The resistive force exerted by air on the airplane has a magnitude of $\mathbf{4.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{}\mathbf{N}$. By Newton’s third law, if the engines exert a force on the exhaust gases to expel them out of the back of the engine, the exhaust gases exert a force on the engines in the direction of the airplane’s travel. This force is called thrust, and the value of the thrust in this situation is $\mathbf{7.50}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{}\mathbf{N}$. (a) Is the work done by the exhaust gases on the airplane during some time interval equal to the change in the airplane’s kinetic energy? Explain. (b) Find the speed of the airplane after it has traveled$\mathbf{5.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{m}$

Short answer

(a) The change in thekinetic energy of the plane is less than the positive work done by the engine thrust. So, themechanical energy does not remain constant for the system.

(b) The speed of the airplane after it has traveled $5.0\times {10}^2\mathrm{m}$is${\mathrm{v}}_{\mathrm{f}}=77.0\text{m/s.}$

Step-by-step solution

Step 1:

In the presence of non-conservative forces, the total change in the mechanical energy of a system should be equal to the work done by the non-conservative forces.

$\sum {\mathbf{W}}_{\mathrm{other}\mathrm{forces}}\mathbf{=}\mathbf{W}\mathbf{=}\mathbf{\Delta K}\mathbf{+}\mathbf{\Delta U}\mathbf{+}{\mathbf{\Delta E}}_{\mathrm{int}}..........(\mathbf{8}\mathbf{.}\mathbf{17})$

Here:

$\mathbf{W}\mathbf{=}$The work by body

$\mathbf{\Delta U}\mathbf{=}$Change in Potential energy

$\mathbf{\Delta K}\mathbf{=}$Change in kinetic energy

${\mathbf{\Delta E}}_{\mathbf{int}}$ Change in internal energy

Given data

Mass oftheairplane:$\mathrm{m}=1.50\times {10}^4\text{\hspace{0.17em}kg}$

Initial velocity oftheairplane:${\mathrm{v}}_{\mathrm{i}}=60.0\text{\hspace{0.17em}m}/\text{s}$

The resistive force due to air:${\mathrm{F}}_{\mathrm{a}}=4.0\times {10}^4\text{\hspace{0.17em}N}$

The engine thrust:${\mathrm{F}}_{\mathrm{E}}=7.5\times {10}^4\text{\hspace{0.17em}N}$

Distance traveled by the plane:$\mathrm{d}=5.0\times {10}^2\text{\hspace{0.17em}m}$

Step 3:

Part (a):

No, the system of the airplane and the surrounding air is non-isolated. Two forces act on the plane during its motion—first is the engine’s thrust, that pushes the plane forward, and air resistance, that acts backward. Since the air resistance force is non-conservative, some of the energy in the system is transformed into internal energy. So, the kinetic energy change of the plane is lesser than the work done by the engine’s thrust. So, the mechanical energy does not remain constant for the system.

Part (b):

Since the plane is in level flight,${\mathrm{U}}_{\mathrm{gf}}={\mathrm{U}}_{\mathrm{gi}}$and the conservation of the energy for non-isolated systems reduces to:

$\begin{array}{c}\mathrm{W}={\mathrm{W}}_{\mathrm{thrust}}\\ ={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}-{\mathrm{f}}_{\mathrm{s}}\\ \mathrm{F}\left(\cos 0^{\mathrm{o}}\right)\mathrm{s}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2-\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2-\mathrm{f}\left(\cos {180}^{\mathrm{o}}\right)\mathrm{s}\end{array}$

This gives:

$\begin{array}{c}{\mathrm{v}}_{\mathrm{f}}=\sqrt{{\mathrm{v}}_{\mathrm{i}}^2-\frac{2(\mathrm{F}-\mathrm{f})\mathrm{s}}{\mathrm{m}}}\\ =\sqrt{{(60.0\text{m/s})}^2+\frac{2[(7.50-4.00)\times {10}^4\text{N}](500\text{m})}{1.50\times {10}^4\text{kg}}}\\ =77.0\text{m/s.}\end{array}$

The final velocity of the airplane is found to be${\mathrm{v}}_{\mathrm{f}}=77.0\text{m/s.}$