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?
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.