A 68.2 kg person jumps from rest off a 2.90 m-high tower straight down into the water. Neglect air resistance during the descent. She comes to rest 1.06 m under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.
The trick is to set h=0 as 1.06 m below the surface. This means the original h is 3.96.
Now mgh = Fd where d is the depth of the water.
To determine the magnitude of the average force that the water exerts on the diver, we can use the concept of work and energy.
First, let's calculate the potential energy of the diver before jumping off the tower:
Potential energy (PE) = mass (m) x gravity (g) x height (h)
PE = 68.2 kg x 9.8 m/s² x 2.9 m
PE = 1899 J
Next, let's calculate the potential energy of the diver when they reach the surface of the water:
PE = mass (m) x gravity (g) x height (h)
PE = 68.2 kg x 9.8 m/s² x 1.06 m
PE = 715 J
Now, subtract the final potential energy from the initial potential energy to find the work done by the water:
Work done (W) = final PE - initial PE
W = 715 J - 1899 J
W = -1184 J
The negative sign indicates that work is done against the direction of motion, which is upward in this case.
Finally, we use the equation for work done:
Work done (W) = force (F) x displacement (d) x cos(θ)
Since the force is acting in the opposite direction to the displacement, the angle between them is 180 degrees. Therefore, cos(180°) = -1.
-1184 J = F x 1.06 m x (-1)
F = -1184 J / (-1.06 m)
F = 1113.2 N
The magnitude of the average force that the water exerts on the diver is approximately 1113.2 N.