# A 7200 lb airplane lands on an aircraft carrier with a speed of 72 ft/s. The plane is caught by...

## Question:

A 7200 lb airplane lands on an aircraft carrier with a speed of 72 ft/s. The plane is caught by an elastic band (k = 998 lb/ft) that has an initial stretch of 5.6 feet. What is the maximum distance the band is stretched?

## Conservation of Energy:

The total energy of a system remains unchanged if the system does not transfer any mass or energy from its surroundings. The increase in one kind of energy of a system occurs at the expense of another type of energy.

Given Data

• The weight of the plane is: {eq}w = 7200\;{\rm{lb}} {/eq}.
• The speed of the plane is: {eq}v = 72\;{\rm{ft/s}} {/eq}.
• The elastic band is: {eq}k = 998\;{\rm{lb/ft}} {/eq}.
• The initial stretch is: {eq}{s_0} = 5.6\;{\rm{ft}} {/eq}.

Let us assume that the maximum distance the band is stretched is {eq}s {/eq}.

The mass of the plane can be calculated as:

$$m = \dfrac{w}{g}$$

The expression for final spring energy is:

$$K{E_f} = \dfrac{1}{2}k{s^2}$$

The expression for the initial spring energy is:

$$K{E_i} = \dfrac{1}{2}k{s_0}^2$$

The expression for the kinetic energy is:

$$KE = \dfrac{1}{2}m{v^2}$$

The expression for the conservation of energy is:

$$K{E_f} = K{E_i} + KE$$

Substitute the known values in the equation:

\begin{align*} \dfrac{1}{2}k{s^2} &= \dfrac{1}{2}k{s_0}^2 + \dfrac{1}{2}m{v^2}\\ \ \\ k{s^2} &= k{s_0}^2 + m{v^2} \end{align*}

Substitute the known values in the equation:

\begin{align*} \left( {998\;{\rm{lb/ft}}} \right){s^2} &= \left( {998\;{\rm{lb/ft}}} \right){\left( {5.6\;{\rm{ft}}} \right)^2} + \left( {\dfrac{{7200\;{\rm{lb}}}}{{32.2\;{\rm{ft/}}{{\rm{s}}^2}}}} \right){\left( {72\;{\rm{ft/s}}} \right)^2}\\ \ \\ {s^2} &= {\left( {5.6\;{\rm{ft}}} \right)^2} + \dfrac{{\left( {\left( {\dfrac{{7200\;{\rm{lb}}}}{{32.2\;{\rm{ft/}}{{\rm{s}}^2}}}} \right){{\left( {72\;{\rm{ft/s}}} \right)}^2}} \right)}}{{998\;{\rm{lb/ft}}}}\\ \ \\ s &= 34.537\;{\rm{ft}} \end{align*}

Thus, the maximum distance the band is stretched is {eq}34.537\;{\rm{ft}} {/eq}.