| Focusing
on the period starting immediatly after the plane begins to move, up to
the point at which the lift force equals the gravity force, the most
apparent conversion of energy is the horizontal acceleration of the
airplane. Since the kinetic energy can only come from converted
mechanical energy in this case, it is a certainty that enough fuel was
burnt to account for the new ground speed of the plane. The kinetic energy of the plane will be equal to .5mv2 . So K= .5*70000kg*51.4m/s2 = 92.46MJ ance But the plane isn't just accelerating itself, it must also overcome the force of friction. The normal force must equal the mass of the plane multiplied by the acceleration due to gravity, So N=9.81m/s2 * 70000kg = 686.7N Using the given value for the co-efficient of friction, we can then determine the force from kinetic friction when the lift is equal to zero Fk=686.7N*.04=27.4N The friction force won't be constant throughout the take off though, because the lift of the plane increases as it goes. Since we are going to assume that this force is changing steadily from 27.4N to 0, the integral will be equal to the average, which is 13.7N. Since work can be expressed as force multiplied by a distance, the energy expended due to friction can easily be calculated by simply multiplying 13.7N by the distance in meters. W=13.7N*5000m=.0685MJ Assuming that the total work done is equal to the change in K plus the work done by friction, we then have a total effort of 92.5MJ. Using the stated value for the energy density of AN-8, 42.8MJ/kg, we have a total fuel burn of 2.161kg. Since the density of AN-8 is this equates to about 3/4 of a gallon of fuel. This is of course a theoretical maximum assuminging a dragless plane. It may seem like a small amount of fuel, but its worth noting that this amount of fuel was burnt over a distance equalling roughly 3 miles. Next |