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