A jet of mass 4,277 kg lands on an airstrip, stopping after a distance of 727 m and has an initial landing speed of 29 m/s.

Question

What is the work done, in Joules, by non-conservative forces to stop the jet?

Answer 1

Assuming a level field, We can solve this from Kinetic Energy only.

Work done on plane by friction = Change in Kinetic Energy = (1/2)mv²

= (1/2) ( 4277 ) (29² )

= 1798478 J = 1.8 MJ, ans

Note that the distance traveled does not matter.

Answer 2

I will set this up for you, but YOU do the calculations.

m = mass = 4277 kg

v₀ = initial speed = 29 m/s

v = final speed = 0 (full stop)

x = stopping distance = 727 m

W = work = ?

I suggest first finding the acceleration a, the force F, and then get the work from

W = Fx

Acceleration:

Use the kinematic relation

v² - v₀² = 2ax

v = 0 ⇒

-v₀² = 2ax

a = -v₀² / (2x)

F = ma = m[-v₀² / (2x)]

W = Fx = m[-v₀² / (2x)] x = -mv₀² / 2

Plug in the values of m and v₀ and get W in joules. The negative sign means the non-conservative force is removing energy from the system.