Question

A Navy jet of mass 10 kg lands on an aircraft carrier and snags a cable to slow it down. If the cable acts like a spring that has a spring constant of 160 N/m and the aircraft is stopped by this spring in 25 m after hooking onto the cable, what is the speed of the jet when it first hooks on to the cable? (assume the jet does no breaking on its own)

Answer 1

Speed of the jet is 100 m/s.

Step-by-step explanation:

Given that,

Mass of the jet, m = 10 kg

Spring constant, K = 160 N/m

Distance covered by the aircraft to stop, x = 25 m

Let v is the landing speed of the plane. It can be calculated using the conservation of mechanical energy. Since, the kinetic energy of the plane must be equal to the elastic potential energy of the spring as :

`(1)/(2)Kx^2=(1)/(2)mv^2`

`v=\sqrt{(Kx^2)/(m)}`

`v=\sqrt{(160* (25)^2)/(10)}`

v = 100 m/s

So, the speed of the jet when it first hooks on to the cable is 100 m/s. Hence, this is the required solution.