How to Calculate Net Force on a Diver Hitting the Water
Introduction
Calculating Speed When Hitting the Water
To calculate the man's speed when he hits the water, we can use the equation: \[ v_f = \sqrt{2gh} \] Where: \( v_f \) = final velocity (speed when hitting the water), \( g \) = acceleration due to gravity (9.81 m/s^2), and \( h \) = height from which the man falls (3.0 m). Substituting the values, we get: \[ v_f = \sqrt{2 \times 9.81 \times 3.0} \] \[ v_f ≈ \sqrt{58.86} \] \[ v_f ≈ 7.67 m/s \] Therefore, the man's speed when he hits the water is approximately 7.67 m/s.Calculating Net Force Inside the Water
To calculate the net force on the man inside the water that brings him to a stop, we can use Newton's second law of motion: \[ F_{\text{net}} = ma \] Where: \( F_{\text{net}} \) = net force, \( m \) = mass of the man (82 kg), and \( a \) = acceleration of the man inside the water (calculated from the time taken to stop). Given that it takes the man 0.55 s to slow down to a stop in the water, we can calculate the acceleration: \[ a = \frac{v - 0}{t} \] Where: \( v \) = final velocity (0 m/s, as he comes to a stop), \( 0 \) = initial velocity, and \( t \) = time taken (0.55 s). Substituting the values, we get: \[ a = \frac{0 - 7.67}{0.55} \] \[ a = -\frac{7.67}{0.55} \] \[ a ≈ -13.94 m/s^2 \] The negative sign indicates deceleration as the man slows down in the water. Now, substituting the acceleration back into Newton's second law equation, we get: \[ F_{\text{net}} = 82 \times -13.94 \] \[ F_{\text{net}} ≈ -1142.28 \] \[ F_{\text{net}} ≈ -1.1 \times 10^3 N \] Therefore, the net force on the man inside the water that brings him to a stop is approximately 1.1x10^3 N upward.Forces on the Man Inside the Water
When the man slows down inside the water, two main forces act on him: buoyancy force and drag force. Buoyancy force acts in the upward direction, opposing the weight of the man, while drag force acts in the opposite direction of the man's motion, slowing him down. The buoyancy force can be calculated using Archimedes' principle: \[ F_{\text{buoyancy}} = \rho \cdot V_{\text{displaced}} \cdot g \] Where: \( \rho \) = density of water, \( V_{\text{displaced}} \) = volume of water displaced by the man, and \( g \) = acceleration due to gravity. The drag force can vary depending on the shape and speed of the diver, but it generally increases with speed. In conclusion, when the man enters the water, various forces act on him, ultimately bringing him to a stop through the net force of 1.1x10^3 N upward.An 82 kg man drops from rest on a diving board 3.0 m above the surface of the water. What is his speed when he hits the water? On hitting the water, it takes him 0.55 s to slow down to a stop. Calculate the net force on him inside the water that brings him to a stop. What are the forces on him when he slows down inside the water?
1. Speed when hitting the water: Approximately 7.67 m/s 2. Net force on him inside the water: Approximately 1.1x10^3 N upward 3. Forces on him when he slows down inside the water: Buoyancy force and drag force