An 70 kg man drops from rest on a diving
board 3.2 m above the surface of the water
and comes to rest 0.57 s after reaching the
water.
The acceleration due to gravity is
9.81 m/s2.
What force does the water exert on the
man? Answer in units of N.
avgforce*time=change in momentum
Calculate the velocity at impact (free fall, 3.2m), multiply by mass.
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To find the force that the water exerts on the man, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, the acceleration is due to the change in velocity as the man falls into the water and comes to rest.
First, we need to find the initial velocity (v0) of the man before hitting the water. We can use the equation of motion:
v = v0 + at
where v is the final velocity (0 m/s) and t is the time taken to reach the water (0.57 s). Rearranging the equation, we find:
v0 = -at
Since the man is in free fall, the acceleration due to gravity acts downwards, so we use a negative sign. Plugging in the values, we get:
v0 = -(9.81 m/s^2)(0.57 s)
v0 = -5.5977 m/s
Now, we can calculate the average force exerted by the water (Fwater) on the man using Newton's second law. The average force is equal to the change in momentum divided by the time interval. The change in momentum is given by the mass of the man (70 kg) multiplied by the change in velocity (v0).
Fwater = (70 kg)(v0 - 0 m/s) / 0.57 s
Fwater = (70 kg)(-5.5977 m/s) / 0.57 s
Now, we can calculate this expression to find the force exerted by the water on the man.