# As a 1.6 \times 10^4 \; kg jet plane lands on an aircraft carrier, its tailhook snags a cable to...

## Question:

As a {eq}1.6 \times 10^4 \; kg {/eq} jet plane lands on an aircraft carrier, its tailhook snags a cable to slow it down. The cable is attached to a spring with spring constant {eq}6.4 \times 10^4 \; N/m {/eq}. If the spring stretches 33 m to stop the plane, what was the plane's landing speed?

## Spring:

Spring is mechanical equipment usually made up of stiff material in such a way that it can store and release the energy that occurred due to deformation. Due to this property, spring is employed as a shock absorber in many vibrating devices.

## Answer and Explanation: 1

**Given Data**

- The mass of the jet is: {eq}{m_j} = 1.6 \times {10^4}\;{\rm{kg}} {/eq}.

- The spring constant of the spring is: {eq}s = 6.4 \times {10^4}\;{\rm{N/m}} {/eq}.

- The stretch in the spring is: {eq}d = 33\;{\rm{m}} {/eq}.

The expression for the kinetic energy of the jet is given by,

{eq}{E_k} = \dfrac{1}{2}{m_j}v_j^2 {/eq}

Here, {eq}{v_j} {/eq} is the speed of the jet.

The expression for the spring energy is given by,

{eq}{E_s} = \dfrac{1}{2}s{d^2} {/eq}

Since, spring is diminishing the speed of jet; the kinetic energy of the jet will be stored by the spring. Therefore, equate the above two energy expressions to calculate the speed of the jet.

{eq}\begin{align*} {E_k} &= {E_s}\\ \dfrac{1}{2}{m_j}v_j^2 &= \dfrac{1}{2}s{d^2}\\ {v_j} &= \sqrt {\dfrac{s}{{{m_j}}}} \times d \end{align*} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} {v_j} &= \sqrt {\dfrac{{6.4 \times {{10}^4}}}{{1.6 \times {{10}^4}}}} \times 33\\ & = 66\;{\rm{m/s}} \end{align*} {/eq}

Thus, the plane's landing speed is {eq}66\;{\rm{m/s}} {/eq}.

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Chapter 17 / Lesson 11In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.