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A jet of mass 4,768 kg lands on an airstrip, stopping after a distance of 414 m. If the jet has...

Question:

A jet of mass 4,768 kg lands on an airstrip, stopping after a distance of 414 m. If the jet has an initial landing speed of 36 m per s, what is the work done, in Joules, by non-conservative forces to stop the jet?

Work & Kinetic Energy:

Change in kinetic energy of an object is always equal to net work done on the object.

Let an object of mass {eq}m {/eq} be moving with initial speed {eq}v_i {/eq} and let its final velocity be {eq}v_f {/eq} after being acted on by a force.

Change in kinetic energy will be given by:

{eq}\displaystyle{\Delta KE = \frac{1}{2}mv^{2}_f - \frac{1}{2}mv^{2}_i} {/eq}.

Therefore, by Work-Energy Theorem, work done on the object by the force can be given by:

{eq}\displaystyle{\color{green}{W = \Delta KE = \frac{1}{2}mv^{2}_f - \frac{1}{2}mv^{2}_i}} {/eq}

Answer and Explanation:

Mass of the jet: {eq}\color{blue}{m = 4768\, \rm kg} {/eq}.

Initial landing speed of the jet: {eq}\color{blue}{v_i = 36\, \rm km/h = 10\, \rm m/s} {/eq}.

Final velocity of the jet: {eq}\color{blue}{v_f = 0\, \rm m/s} {/eq}.

Change in kinetic energy is given by:

{eq}\displaystyle{\begin{align*} \color{blue}{\Delta KE} &= \color{blue}{\frac{1}{2}mv^{2}_f - \frac{1}{2}mv^{2}_i}\\ &= \color{blue}{\frac{1}{2}\times 4768\times 0^{2} - \frac{1}{2}\times 4768 \times 10^{2}}\\ &= \color{blue}{-238,400\, \rm J}\\ \end{align*}} {/eq}

By Work-Energy Theorem, net work done on the jet by non-conventional energy is given by:

{eq}\displaystyle{\begin{align*} \color{red}{W} &= \color{blue}{\Delta KE}\\ &= \color{red}{-238,400\, \rm J}\\ \end{align*}} {/eq}


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Practice Applying Work & Kinetic Energy Formulas

from Physics 101: Help and Review

Chapter 17 / Lesson 4
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