A skydiver of mass 100 kg experiences air resistance of 500 N, and an acceleration of about?

Answer 1

There are two forces acting on the sky diver....one force is explicitly stated (air resistance) and the other is implied (the sky diver's weight).

Weight = mass * gravity
Where gravity is the gravitational acceleration experienced by the object (9.81 m/s^2).
W = 100 kg * 9.81 m/s^2
W = 981 Newtons

In addition to weight, you are given the force due to air resistance as 500 Newtons.

Now visualize what is happening (draw a force diagram).
The force due to gravity is pulling the sky diver down towards the ground. The force due to air resistance is pushing the sky diver in the opposite direction of motion, so since he is falling down the air resistance is pushing up.
The forces of weight and air resistance are acting in opposite directions so to add these two force vectors you need to subtract their magnitudes.

If we call the positive direction "down", then there is a positive force of 981 N acting on the object, but from this we subtract the air resistance's force (adding a negative force is the same as subtracting a positive force). So we subtract off 500 N from the 981 N and we are left with the overall net force of 481 Newtons.

Now that we know the net force acting on the sky diver (481 N), we can use Newton's 2nd law to find his acceleration.
F = ma
Where F is the net force, m is the sky diver's mass, and a is the acceleration the sky diver experiences.
481 N = 100 kg * a
Solving for the acceleration,
a = 481 N / 100 kg
a = 4.81 m/s^2

So the net acceleration of the sky diver is 4.81 m/s^2 in the positive direction...which we had previously defined to be downward. The sky diver accelerates down towards the ground at a rate of 4.81 m/s^2.

Answer 2

7.98 sum of forces problem. F1 (gravity) = ma = 539 F2 (air density resistance) = -100 (going in opposite direction) F1 + F2 = sumF = 439 going back to acceleration equation: sumF = ma, 439/55 = 7.98, slightly less than gravity which is good.

Answer 3

F = Mass X Acceleration.

Just make sure your have your dimensions correct.

probably meters per second squared. rather than feet per second squared

No, I'm not going to do it for you. you have to do your homework cause this one is going to be on the test.
Sincerely:
your teacher.

Answer 4

32 feet per second till he reaches max velocity of 120 mph ??