A 66 kg diver steps off a 13 m tower and drops, from rest, straight down into the water. If he comes to rest 4.4 m beneath the surface, determine the average resistance force exerted on him by the water.
force*distance=mgh Now there is a trickher, h is not 13m he actually fell 13+4.4m, but I think here I would ignore it, because if you include that, you need to include the force of bouyancy, which you have little info to do.
To determine the average resistance force exerted on the diver by the water, we can use the principle of work and energy. We'll need to calculate the work done by the water, and then divide it by the distance traveled by the diver.
1. First, let's find the potential energy of the diver at the top of the tower. The potential energy (PE) is given by the formula PE = mgh, where m is the mass (66 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the tower (13 m).
PE = (66 kg)(9.8 m/s^2)(13 m) = 8,796 J
2. Next, let's find the speed of the diver when he reaches a depth of 4.4 m. We can use the principle of conservation of energy, neglecting any energy losses due to air resistance or other factors. The initial potential energy is converted entirely into kinetic energy and the energy required to overcome the resistance force.
PE = KE + W, where KE is the kinetic energy and W is the work done against the resistance force.
KE = 1/2mv^2, where v is the final velocity of the diver.
The work done against the resistance force is equal to the resistance force multiplied by the distance traveled.
W = Fr, where F is the resistance force and r is the distance traveled (13 m + 4.4 m = 17.4 m).
Plugging in the values, we get: 8,796 J = 1/2(66 kg)v^2 + Fr.
3. We also know that the final velocity, when the diver reaches a depth of 4.4 m, is 0 m/s since the diver comes to rest.
Therefore, 8,796 J = 0 + F(17.4 m).
4. Solving for the resistance force, F: 8,796 J = 17.4 F.
F = 8,796 J / 17.4 m.
F ≈ 505.75 N.
Therefore, the average resistance force exerted on the diver by the water is approximately 505.75 Newtons.