Answer:
[tex]F_o = 890.67 N[/tex]
Explanation:
initial speed of the aircraft = 0
it covers a distance = 100 m
time taken by it = 53 s
so we will have
[tex]d = v_i t + \frac{1}{2}at^2[/tex]
[tex]100 = 0 + \frac{1}{2}a(53^2)[/tex]
[tex]a = 0.071 m/s^2[/tex]
now we know that there are two forces on the aircraft
1) applied force in forward direction
2) friction force in opposite direction
so we will have
[tex]F - \mu mg = ma[/tex]
[tex]F - (0.02)(200,000)(9.81) = 200,000(0.071)[/tex]
[tex]F = 39,240 + 14,200[/tex]
[tex]F = 53440 N[/tex]
so net force by 60 officers is 53440 N
so force due to each officer is given as
[tex]F_o = \frac{53440}{60} = 890.67 N[/tex]