stunt pilot weighing 0.70 kN performs a vertical circular dive of radius 0.80 km.At the bottom of the dive, the pilot has a speed of 0.20 km/s which at that instant is notchanging. What force does the plane exert on the pilot

Respuesta :

To solve this problem it is necessary to apply the concepts of the Centripetal Force and the force caused by gravity. The centripetal force can be described as

[tex]F_c = \frac{mv^2}{r}[/tex]

Where,

m = Mass

v = Velocity

r = Radius

At the same time the force caused by the weight can be described as

[tex]F_g = mg \rightarrow m = \frac{F_g}{g}[/tex]

Where,

m = mass

g = Gravity

If we make a sum of Forces, the forces that act vertically on the body, both in the upward and downward direction must be equivalent to the centripetal Force, therefore

[tex]F_f - F_g = F_c[/tex]

Here [tex]F_f[/tex] represents the force from Plane, then:

[tex]F_f - F_g =\frac{mv^2}{r}[/tex]

If we put the mass of the body according to the weight we would have to:

[tex]F_f - F_g = \frac{(\frac{F_g}{g})v^2}{r}[/tex]

[tex]F_f -7kN = \frac{(\frac{7kN}{9.8})(0.2)^2}{0.8}[/tex]

Converting to SI:

[tex]F_f - 7000 =\frac{(\frac{7000N}{9.8})(200)^2}{800}[/tex]

[tex]F_f = 4271N[/tex]

Therefore the forces that the plane exert on the pilot is 4271N