Respuesta :
As per the question the mass of two falling sky drivers is 132 kg.
First we have to calculate their acceleration.
Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.
The earth pulls the object with a force equal to the weight of the body.
Hence the force gravity F=W= mg [ here m is mass of the body]
Here m =132 kg.
Hence force of gravity F= mg
=132 kg ×9.8 m/s^2
=1293.6 kg m/s^2
=1293.6 N [ here N[newton] is the unit of force.]
As per the question the air resistance is one fourth of weight of the bodies.
Hence air resistance F' =1/4 mg
[tex]=\frac{1}{4} *1293.6N[/tex]
[tex]=323.4 N[/tex]
Here F acts in vertically downward direction while F' acts in vertically upward direction.
Hence the net force acting on the particle is F-F'.
[tex]F_{net} =1293.6N -323.4N[/tex]
[tex]=970.2 N[/tex]
From Newton's second law of motion we know that net force is the product of mass and acceleration i.e
[tex]F_{net} =ma[/tex] [Here a is the acceleration]
[tex]a =\frac{F_{net} }{m}[/tex]
[tex]= \frac{970.2}{132} m/s^2[/tex]
[tex]=7.35 m/s^2[/tex]
In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.
we know that F= ma
=m×0
=0 N
Hence they will not get any force when they will descend with a uniform speed.
(a) The acceleration of two falling sky divers when the upward force of air resistance is equal to one-fourth of their weight is 7.35 m/s².
(b) When the divers descend at constant speed, the air resistance is 1293.6 N.
The given parameters;
- mass of the skydivers, m = 132 kg
- air resistance, R = ¹/₄ of their weight
The weight of the skydivers;
W = mg
W = 132 x 9.8
W = 1293.6 N
The air resistance on the skydivers is calculated as follows;
R = 0.25 x 1293.6
R = 323.4 N
The resultant force of the skydivers is calculated as follows;
∑F = 1293.6 N - 323.4 N = ma
970.2 = ma
[tex]a = \frac{F}{m} \\\\a = \frac{970.2}{132} \\\\a = 7.35 \ m/s^2[/tex]
When the divers descend at constant speed, the acceleration of the divers is zero.
W - R = ma
1293.6 - R = 132(0)
1293.6 - R = 0
R = 1293.6 N
Thus, when the divers descend at constant speed, the air resistance is 1293.6 N.
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