# Forced landing of a plane - Find the minimum length of the barage

## Homework Statement

A 1000 kg plane is trying to make a forced landing on the deck of a 2000kg barage at rest on the surface of a calm sea. The only frictional force to consider is between the plane's wheels and the deck, and this braking force is constant and equal to one-quater of the plane's weight. What must the minimum length of the barage be, in order that the plane can stop safetly on the deck, if the plane touches down just at the rear end of the deck with a velocity of 50m/s towards the front of the barage?

## The Attempt at a Solution

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cristo
Staff Emeritus
You need to show some work before we can help. How about trying to fill in numbers 2 and 3 of the above template?

If the barge is allowed to move when the plane touches down - are we to assume it moves without 'friction' through the water ?

i have no clue abt what i have to do :(

i did tried solving the questions and I did it the following way:

Pi=Pf
(Mc x Vc)+(Mc x Vc)= (Mc x Vl) x Vcl
30 000kgm/s /2000 kg = Vcl

15m/s = Vcl

Ei = Ef
kinetic enerygy = mgh
0.5(2000)(15 x 15) = ( 2000 x 9.8 m/s) h
(225000 kgm/s2) / (19600 kgm/s2)
= 11.5 kgm/s2

plz let me no if its right or wrong...thank you

To calculate the stopping distance you can use energy conservation.
The stopping is done by friction, and the work done by friction is force*distance.

This work is equivalent to the kinetic energy of the plane when it touches down. Equate the work and energy and you can compute the distance. You can also find the time it takes to stop.

I'm ignoring any movement of the barge. If the barge moves ( and I think it will) then the distance above can be reduced by the amount the barge moves from the touchdown to the plane stopping. This distance is given by 1/2at^2 where a = f/M,
f = mu.m.g and M and m are the masses of the barge and plane and t is the time to stop.

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