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
[tex]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[/tex]
Force is given by
[tex]F=ma+f\\\Rightarrow F=74000\times 0.198871271164+14000\\\Rightarrow F=28716.4740661\ N[/tex]
The force with which the team pulls the plane is 28716.4740661 N
[tex]v=u+at\\\Rightarrow v=0+0.198871271164\times 6.1\\\Rightarrow v=1.2131147541\ m/s[/tex]
The speed of the plane is 1.2131147541 m/s
Kinetic energy is given by
[tex]K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}\times 74000\times 1.2131147541^2\\\Rightarrow K=54450.9540448\ J[/tex]
Work done is given by
[tex]W=Fs\\\Rightarrow W=28716.4740661\times 3.7\\\Rightarrow W=106250.954045\ J[/tex]
The fraction is given by
[tex]\dfrac{54450.9540448}{106250.954045}=0.512474965841[/tex]
The teams 51.2474965841% of the work goes to kinetic energy of the plane.
Given values are:
(a)
By using the relation,
→ [tex]s = ut+\frac{1}{2} at^2[/tex]
[tex]3.7=0\times 6.1+\frac{1}{2}\times a\times 6.1^2[/tex]
[tex]a = \frac{3.7\times 2}{6.1^2}[/tex]
[tex]= 0.19888 \ m/s^2[/tex]
now,
The force will be:
→ [tex]F = ma+f[/tex]
[tex]= 7400\times 0.19888+14000[/tex]
[tex]= 28716.48 \ N[/tex]
(b)
The speed of plane,
→ [tex]v = u+at[/tex]
[tex]= 0+0.19888\times 6.1[/tex]
[tex]= 1.214 \ m/s[/tex]
(c)
The Kinetic energy is:
→ [tex]K = \frac{1}{2}mv^2[/tex]
[tex]= \frac{1}{2}\times 74000\times 1.214[/tex]
[tex]= 54450.95 \ J[/tex]
now,
The work done will be:
→ [tex]W = Fs[/tex]
[tex]= 28716.48\times 3.7[/tex]
[tex]= 1060250.96 \ J[/tex]
hence,
The fraction will be:
= [tex]\frac{54450.96}{106250.96}[/tex]
= [tex]0.512474[/tex]
or,
= [tex]51.25[/tex] (%)
Thus the responses above are correct.
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