Question

As a 15000 kg jet plane lands on an aircraft carrier, its tail hook snags a cable to slow it down. The cable is attached to a spring with spring constant 60000 N/m.If the spring stretches 30 m to stop the plane, what was the plane's landing speed?

Answer 1

Answer:

60 m/s

Explanation:

From the law of conservation of energy,

The kinetic energy of the plane = Energy of store in the spring when the plane lands.

1/2mv²  = 1/2ke²

making v the subject of the equation.

v = √(ke²/m).................... Equation 1

Where  v = the plane landing speed, k = spring constant, e = extension. m = mass of the plane.

Given: m = 15000 kg, k = 60000 N/m, e = 30 m.

Substitute into equation 1

v = √(60000×30²/15000)

v = √(4×900)

v = √(3600)

v = 60 m/s.

Hence the plane's landing speed = 60 m/s

Answer 2

The landing speed of plane is 328.63 m/s.

Given data:

The mass of jet plane is, m = 15000 kg.

The value of spring constant is, k = 60000 N/m.

The value of stretching distance is, x = 30 m.

Here we can apply the concept of work done by the spring force, which is equivalent to kinetic energy change as per the work-energy principle. Then,

$W = \Delta KE\\\\F \times x = \dfrac{1}{2}mv^{2}$

Here, v is the plane's landing speed.

Solving as,

$\dfrac{1}{2} \times k \times x^{2} \times x = \dfrac{1}{2}mv^{2}\\\\ k \times x^{3} = m \times v^{2}\\\\\v = \sqrt{\dfrac{60000\times 30^{3}}{15000}}\\\\\\v = 328.63 \;\rm m/s$

Thus, we can conclude that the landing speed of plane is 328.63 m/s.

Learn more about the Spring force here:

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