A 62.0 kg diver falls from rest into a swimming pool from a height of 3.50 m. It takes 1.92 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
The impluse given to the diver is equal to the momentum change.
First, find the velocity at impact.
vf=sqrt(2*g*h)
Then, in the water,
Forceaverage*time=mass*changeinvelocity.
yeah you were right thanks.
To find the magnitude of the average force exerted on the diver, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the product of its mass and acceleration:
F = m * a
In this case, the force we want to find is the average force, the mass of the diver is given as 62.0 kg, and the acceleration can be calculated using the kinematic equation:
v = u + a * t
where:
v = final velocity (0 m/s as the diver stops)
u = initial velocity (0 m/s as the diver falls from rest)
a = acceleration
t = time taken to stop
Rearranging the equation, we get:
a = (v - u) / t
Since we know the final velocity (0 m/s), initial velocity (0 m/s), and the time taken to stop (1.92 s), we can calculate the acceleration.
a = (0 - 0) / 1.92
a = 0
Now, we have the mass of the diver (62.0 kg) and the acceleration (0 m/s^2). Substituting these values into Newton's second law:
F = m * a
F = 62.0 kg * 0 m/s^2
F = 0 N
Therefore, the magnitude of the average force exerted on the diver during that time is 0 N.