Question

The Special Olympics raises money through "plane pull" events in which teams of 25 people compete to see who can pull a 74,000 kg airplane 3.7 m across the tarmac. The inertia of the plane is an issue--but so is the 14,000 N rolling friction force that works against the teams. If a team pulls with a constant force, and thus a constant acceleration, and moves the plane 3.7 m in 6.1 s (an excellent time),a. What force does the team pull with?b. What is the speed of the plane at the end of the pull?c. What fraction of the teamâs work goes to kinetic energy of the plane?

Answer 1

Answer:

28716.4740661 N

1.2131147541 m/s

51.2474965841%

Explanation:

m = Mass of plane = 74000 kg

s = Displacement = 3.7 m

f = Frictional force = 14000 N

t = Time taken = 6.1 s

u = Initial velocity = 0

v = Final velocity

$s=ut+\frac{1}{2}at^2\\\Rightarrow 3.7=0\times 6.1+\frac{1}{2}\times a\times 6.1^2\\\Rightarrow a=\frac{3.7\times 2}{6.1^2}\\\Rightarrow a=0.198871271164\ m/s^2$

Force is given by

$F=ma+f\\\Rightarrow F=74000\times 0.198871271164+14000\\\Rightarrow F=28716.4740661\ N$

The force with which the team pulls the plane is 28716.4740661 N

$v=u+at\\\Rightarrow v=0+0.198871271164\times 6.1\\\Rightarrow v=1.2131147541\ m/s$

The speed of the plane is 1.2131147541 m/s

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}\times 74000\times 1.2131147541^2\\\Rightarrow K=54450.9540448\ J$

Work done is given by

$W=Fs\\\Rightarrow W=28716.4740661\times 3.7\\\Rightarrow W=106250.954045\ J$

The fraction is given by

$\dfrac{54450.9540448}{106250.954045}=0.512474965841$

The teams 51.2474965841% of the work goes to kinetic energy of the plane.