Question

A small plane that moves along a 1.4 kmkm long runway during take-off accelerates from rest to a speed of 10.4 m/s in 4.49 minutes . If the mass of the plane is 870 kgkg , what is the total work done on the plane?

Answer 1

There are three equations of motion given by

1. v = u+ at

2. v2 - u2 = 2aS

3. S = ut+at2/2

where S is the distance covered, v and u are the final and initial velocities respecitively and t is the time taken.

We are to calculate the work done on the plane, and we know that expression for work done (W) is given by

W = FS ..............eqn 1

where F is the force applied and S is the change in displacement of the  body on the application of the force.

Since, F = ma

where m is the mass of the body and a is the acceleration attained, using this expression for F in eqn 1

we get

W = maS ............eqn 2

Now, we have m =870 kg

S = 1.4 km = 1.4*103 m ( 1 km = 103 m )

We are to calculate the value of acceleration a in order to calculate the work done on the plane.

According to second eqn of motion, we have

v = u + at ................ eqn 3

Since, it is given that plane started from the rest. Hence, u = 0

also, v = 10.4 m /s  

Time taken, t = 4.49 min = 4.49*60 = 269.4 s ( 1 min = 60 seconds)

Using value of v, u, t in eqn 3, we get

10.4 = 0+ a*269.4

10.4 = a*269.4

10.4/269.4 = a

0.0386 = a

a = 0.0386 m/s2

i.e acceleration attained by plane in given time = 0.0386 m/s2

Using this value of a, m , S in eqn 2, we get

W = 870*0.0386*1.4*103 = 47.014 J

Hence, work done on the plane is = 47.014 J