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Chapter 6: Problem 59

A 67.0-kg person jumps from rest off a 3.00-m-high tower straight down into the water. Neglect air resistance. She comes to rest 1.10 m under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is non conservative.

Short Answer

Expert verified
The average force exerted by the water is approximately 1792.58 N.

Step by step solution

01

Calculate the Diver's Initial Velocity

The diver jumps from a height of 3.00 m. We can use the kinematic equation to find her velocity just before entering the water: \( v^2 = u^2 + 2gh \) where \( u = 0 \) (initial velocity), \( g = 9.8\, \text{m/s}^2 \) (acceleration due to gravity), and \( h = 3.00\, \text{m} \). The velocity \( v \) when she hits the water is calculated as follows:\[ v = \sqrt{0 + 2(9.8)(3.00)} \]\[ v = \sqrt{58.8} \approx 7.67\, \text{m/s} \]
02

Calculate the Diver's Change in Momentum

Once in the water, the diver comes to rest, indicating her final velocity is \( 0 \). Her initial velocity when she enters the water is \( 7.67\, \text{m/s} \). Using her mass, \( m = 67.0\, \text{kg} \), the change in momentum \( \Delta p \) is:\[ \Delta p = m \times (v_f - v_i) = 67.0 \times (0 - 7.67) \approx -514.89\, \text{kg} \cdot \text{m/s} \]
03

Calculate the Deceleration Under Water

The diver comes to rest after traveling 1.10 m in the water. Using the equation \( v_f^2 = v_i^2 + 2as \) and solving for \( a \) where \( v_f = 0 \), \( v_i = 7.67 \), and \( s = 1.10 \), we have:\[ 0 = (7.67)^2 + 2a(1.10) \]\[ a = -\frac{(7.67)^2}{2 \times 1.10} \approx -26.74\, \text{m/s}^2 \]
04

Determine the Average Force Exerted by the Water

The average force exerted by the water can be found using Newton's second law: \( F = m \times a \). With \( m = 67.0\, \text{kg} \) and \( a = -26.74 \), the force \( F \) is:\[ F = 67.0 \times 26.74 \approx 1792.58\, \text{N} \]Note that the force is positive as it is exerted upwards against the direction of motion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

kinematic equations
Kinematic equations are a set of equations that describe the motion of objects under constant acceleration. They help predict future motion based on initial conditions and forces. In our exercise, the diver begins at rest before jumping off a 3.00-meter tower. By using the kinematic equation, \( v^2 = u^2 + 2gh \), we can determine her velocity just before she hits the water.
This equation works because it accounts for the acceleration due to gravity \( (g = 9.8 \text{ m/s}^2) \) and the initial velocity \( (u = 0 \text{ m/s}) \). Applying the numbers, the diver's velocity just before impact with water comes to approximately 7.67 m/s.
momentum change
Momentum is a product of an object's mass and velocity. Change in momentum occurs when an object's speed changes, which usually happens when a force is applied. For the diver, her momentum changes as she goes from moving to resting in the water.
Initially, her momentum is non-zero due to her speed upon striking the water. With an initial velocity of 7.67 m/s and mass of 67.0 kg, her change in momentum becomes \(-514.89\, \text{kg} \cdot \text{m/s}\). This reduction in momentum indicates a considerable force exerted by the water to stop the diver.
deceleration
Deceleration is the reduction of speed or velocity. In this problem, it's the process of the diver slowing down upon entering the water. It is essentially negative acceleration and is determined by using the kinematic equation again.
Here, we know her initial velocity (7.67 m/s), final velocity (0 m/s after stopping), and the distance over which she decelerates (1.10 m). By substituting these values into the equation \( v_f^2 = v_i^2 + 2as \), we can calculate the deceleration as approximately -26.74 m/s虏. The negative sign indicates a reduction in speed in the opposite direction of her entering velocity.
Newton's second law
Newton's second law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (\( F = ma \)). It's a key principle in understanding how forces result in changes in motion.
In the context of the diver, once we know the mass (67.0 kg) and the deceleration (-26.74 m/s虏), we can calculate the force the water applies to stop her. This calculation reveals an average force of approximately 1792.58 N. Importantly, this force is directed upwards as the water resists her downwards motion, allowing her to come to a complete stop.

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Most popular questions from this chapter

A slingshot fires a pebble from the top of a building at a speed of 14.0 m/s. The building is 31.0 m tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of 22.0 m. She pushes in a direction \(29.0^{\circ}\) below the horizontal. A 48.0-N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

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The (non conservative) force propelling a \(1.50 \times 10^{3}-kg\) car up a mountain road does \(4.70 \times 10^{6} J\) of work on the car. The car starts from rest at sea level and has a speed of 27.0 \(m /s\) sat an altitude of \(2.00 \times 10^{2} m\) above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are non conservative forces.

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