A light plane (mass = M) makes an emergency landing on a short runway. With its

Question

A light plane (mass = M) makes an emergency landing on a short runway. With its engine off, it lands at speed v0. A hook on the plane snags a cable attached to a sandbag (mass = m) and drags the sandbag along. The coefficient of friction between the sandbag and the runway is ?, and the plane's brakes give a retarding force of Fb. How far will the plane go before it stops?
Data: v0 = 37.0 m/s; M = 994 kg; m = 104 kg; ? = 0.37; Fb = 1575 N.

Since it's an inelastic collision I used conservation of momentum to solve for the new velocity after the plane grabs the bag.

994*37 = (994+104) v

v = 33.495 m/s

The kinetic energy of the plane = 1/2(994)((37^2) = 680393 J
The kinetic energy of the sand bag+plane = 1/2(994+104)*33.495^2 = 615931.35 J

So the energy lost is 64,461.65 J

I also know the total retardant force = Fb+Force of Friction = 1575+umg = 1952.104 N

How do you proceed from here?

Explanation / Answer

using omentum conservation we have M*Vo = (m+M)*v and using energy conservation we have 1/2*(M+m)*v^2 = x*[uo*m*g + Fb] where x is the required distance

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