An 82 kg man drops from rest on a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What is the net force on the diver as he is brought to rest?

Answer 1

First, we need to calculate the velocity of the diver right before he hits the water. We can use the equation of motion:

v = u + at

where:
- v is the final velocity (0 m/s since he comes to rest)
- u is the initial velocity (0 m/s since he drops from rest)
- a is the acceleration (which we need to calculate)
- t is the time taken to come to rest (0.55 s)

Rearranging the equation, we get:

a = (v - u) / t
a = (0 - 0) / 0.55
a = 0 m/s^2

Now, we can calculate the force using Newton's second law:

F = ma

where:
- F is the net force
- m is the mass of the diver (82 kg)
- a is the acceleration (0 m/s^2)

Plugging in the values, we get:

F = 82 kg * 0 m/s^2
F = 0 N

Therefore, the net force on the diver as he is brought to rest is 0 N.