Chapter 6, Problem 56P

Part (a)

To determine

Landing speed in the absence of air resistance.

Answer to Problem 56P

Solution:

1100 km/h

Explanation of Solution

Given:

Mass of flight including pilot $\text{=980}\text{kg}$

Initial horizontal speed of flight $\text{=480}\text{km/h}$

Initial height of pilot $\text{=3500}\text{m}$

Final land speed of flight $\text{=210}\text{km/h}$

Formula used: Conservation of energy (only gravitational force is acting which is conservative in nature)

  $K.E_1+P.E_1=K.E_2+P.E_2$

  $W_{NC}=\Delta K.E+\Delta P.E$

  $W_{NC}$: Work done by non-conservative force.Here, it is air resistance.

Calculation:

  $\begin{array}{l}{v}_1\text{=480 km/h}\\ \text{ =}\frac{\text{480×1000}}{\text{60×60}}\text{m/s}\\ \text{ =133}\text{.33 m/s}\end{array}$

Initial kinetic energy:

  $\begin{array}{c}\left(K.E_1\right)=\frac{1}{2}m{v}_1{}^2\\ \text{ =}\frac{\text{1}}{\text{2}}\text{×980}\times {\left(133.3\right)}^{\text{2}}\end{array}$

Initial potential energy:

  $\begin{array}{c}\left(P.E_1\right)=mg{h}_1\\ =9\text{80×9}\text{.80×3500 J }\end{array}$

Final kinetic energy:

  $\begin{array}{c}\left(K.E_2\right)=\frac{1}{2}m{v}_2{}^2\\ =\frac{1}{2}\times m\times v_2{}^2\end{array}$

Final potential energy:

  $\begin{array}{c}\left(P.E_2\right)=mg{h}_2\\ =m\text{×g×}\left(\text{0}\right)\\ \text{=}\text{0}\end{array}$

  $\text{K}{\text{.E}}_{\text{1}}\text{+P}{\text{.E}}_{\text{1}}\text{=K}{\text{.E}}_{\text{2}}\text{+P}{\text{.E}}_{\text{2}}$

  $\begin{array}{l}\frac{1}{2}\times \left(980\right)\times {\left(133.33\right)}^2+980\times 9.80\times 3500=\frac{1}{2}\left(980\right)v_2{}^2\\ v_2=\sqrt{{\left(133.33\right)}^2+2\times 9.80\times 3500}\\ =293.89\text{m/s}\\ =\frac{293.89}{1000}\times 3600\text{km/h}\\ =1058.004\text{km/h}\\ \approx \text{1100 km/h}\end{array}$

Conclusion:Landing speed in the absence of air resistance would be 1100 km/h.

Part (b)

To determine

The average force of air resistance exerted if it came in at a constant glide angle of $12°$ to the earth’s surface.

Answer to Problem 56P

Solution:2400 N

Explanation of Solution

Given:

Mass of flight including pilot $\text{=980}\text{kg}$

Initial horizontal speed of flight $\text{=480}\text{km/h}$

Initial height of pilot $\text{=3500}\text{m}$

Final land speed of flight $\text{=210}\text{km/h}$

Formula used: Conservation of energy (only gravitational force is acting which is conservative in nature)

  $K.E_1+P.E_1=K.E_2+P.E_2$

  $W_{NC}=\Delta K.E+\Delta P.E$

  $W_{NC}$: Work done by non-conservative force here it is air resistance.

Calculation:

If air resistance is acting, then speed of plane with which it landed is

  $\begin{array}{l}\text{=}\text{210}\text{km/h}\\ \text{=}\frac{\text{210×1000}}{\text{60×60}}\text{m/s}\\ \text{=}\text{58}\text{.33}\text{m/s}\end{array}$

Kinetic energy in this case $\text{=}\frac{\text{1}}{\text{2}}\text{×980×}{\left(\text{58}\text{.33}\right)}^{\text{2}}\text{J}$

If air resistance does not act, then speed with which flight land at a speed $\text{=293}\text{.89}\text{m/s}$

Kinetic energy in this case $\text{=}\frac{\text{1}}{\text{2}}\text{×980×}{\left(293.89\right)}^{\text{2}}\text{J}$

Work done by air resistance (magnitude only)

  $\text{=}\frac{\text{1}}{\text{2}}\text{×980}\left[{\left(293.89\right)}^2-{\left(\text{58}\text{.33}\right)}^{\text{2}}\right]\text{J}$

Work done by air resistance can also find out

= air resistance $\times$ distance travelled

Distance travelled can be find out

  $\begin{array}{l}\sin \theta =\frac{p}{h}\\ \sin 12°=\frac{3500}{h}\\ h=\frac{3500}{\sin 12°}\end{array}$

Air resistance:

  $\begin{array}{c}f=\frac{\text{work done by air resistance}}{\text{distance travelled}}\\ \text{ =}\frac{\frac{\text{1}}{\text{2}}\text{×980}\times \left[{\left(\text{293}\text{.89}\right)}^{\text{2}}\text{-}{\left(\text{58}\text{.33}\right)}^{\text{2}}\right]}{\frac{\text{3500}}{\text{sin12}°}}\\ \text{=2415}\text{.02}\text{N}\\ \approx \text{2400}\text{N}\end{array}$

Conclusion: Average force of friction acting on the flights $\text{=2400N}$