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Chapter 7: Problem 2

At provides a review of the concepts that are involved in this problem. A \(62.0-\mathrm{kg}\) person, standing on a diving board, dives straight down into the water. Just before striking the water, her speed is \(5.50 \mathrm{~m} / \mathrm{s}\). At a time of \(1.65 \mathrm{~s}\) after she enters the water, her speed is reduced to \(1.10 \mathrm{~m} / \mathrm{s}\). What is the net average force (magnitude and direction) that acts on her when she is in the water?

Short Answer

Expert verified
The net average force is approximately 165.54 N upward.

Step by step solution

01

Define Known Variables

We know the initial mass of the person, \( m = 62.0 \text{ kg} \), her initial velocity \( v_i = 5.50 \text{ m/s} \), her final velocity \( v_f = 1.10 \text{ m/s} \), and the time interval \( t = 1.65 \text{ s} \).
02

Calculate the Change in Velocity

The change in velocity \( \Delta v \) can be found using the formula \( \Delta v = v_f - v_i \). Substituting the known values, we get \( \Delta v = 1.10 \text{ m/s} - 5.50 \text{ m/s} = -4.40 \text{ m/s} \).
03

Calculate the Average Acceleration

Use the formula for acceleration \( a = \frac{\Delta v}{t} \). Substituting the values, \( a = \frac{-4.40 \text{ m/s}}{1.65 \text{ s}} \approx -2.67 \text{ m/s}^2 \). This negative sign indicates deceleration.
04

Apply Newton's Second Law to Find Net Force

Newton's second law states \( F = ma \). Using this formula, substitute \( m = 62.0 \text{ kg} \) and \( a \approx -2.67 \text{ m/s}^2 \) to find \( F \). Thus, \( F = 62.0 \text{ kg} \times -2.67 \text{ m/s}^2 \approx -165.54 \text{ N} \).
05

Determine the Direction of the Force

The net average force is negative, indicating that it acts in the upward direction, opposite to the direction of motion.

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Key Concepts

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

Deceleration
When an object's speed decreases, this is termed deceleration. It essentially means slowing down and occurs when the acceleration is in the opposite direction to the velocity. In this exercise, a diver hits the water at a certain speed and slows down as they continue downwards through the water.
Here is how deceleration works:
  • It is represented as a negative acceleration because the object's speed is decreasing over time.
  • The diver's original speed was higher compared to her speed later, indicating a slowdown.
  • In equations, if the change in velocity comes out negative, it signifies deceleration.
This means that when we calculate the change in the diver's speed, taking the final speed minus the initial speed, a negative result shows she is decelerating. Importantly, deceleration is common in everyday scenarios, such as vehicles stopping, sports athletes slowing down, or any object resisting a force that opposes motion.
Net Force Calculation
To better understand the forces involved in the diver's motion, we use Newton's Second Law. This law connects force, mass, and acceleration, forming the equation: \[ F = ma \]
This means the force applied to an object is equal to its mass times its acceleration.
  • In the diver's case, the mass is the constant weight of the person, which is 62 kg.
  • The acceleration is her deceleration, a change of \[ -2.67 \, \text{m/s}^2 \]
  • Net force (\[ F \]) is calculated by multiplying 62 kg and \[ -2.67 \, \text{m/s}^2 \], resulting in \[ -165.54 \, \text{N} \].
The negative sign of the force is noteworthy. It indicates that the force is acting against the direction of the dive, essentially acting upwards in the opposite direction to motion. This scenario is important in understanding real-world applications like stopping distances and how external forces need to overcome acceleration to create deceleration.
Velocity Change
Velocity tells us how fast something is moving and in what direction. When a person's velocity changes, it implies they are either speeding up or slowing down.The main idea is:
  • The diver starts at 5.5 m/s and ends at 1.1 m/s, showing a clear decrease.
  • The formula used here is:\[ \Delta v = v_f - v_i \]
  • Plugging in the values gives:\[ 1.1 \, \text{m/s} - 5.5 \, \text{m/s} = -4.4 \, \text{m/s} \]
The negative result indicates a decrease in speed, or slowing down. Identifying the velocity change is essential because it directly relates to both the deceleration and the net force experienced by the diver.
Understanding velocity change is crucial not just in this context but also in analyzing motion in general. Whether it鈥檚 how quickly a car slows to a halt or the rate at which an athlete reduces speed, this concept helps define and solve many real-world problems.

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Most popular questions from this chapter

At illustrates how to model a problem similar to this one. An automobile has a mass of \(2100 \mathrm{~kg}\) and a velocity of \(+17 \mathrm{~m} / \mathrm{s}\). It makes a rear-end collision with a stationary car whose mass is \(1900 \mathrm{~kg} .\) The cars lock bumpers and skid off together with the wheels locked. (a) What is the velocity of the two cars just after the collision? (b) Find the impulse (magnitude and direction) that acts on the skidding cars from just after the collision until they come to a halt. (c) If the coefficient of kinetic friction between the wheels of the cars and the pavement is \(\mu_{k}=0.68,\) determine how far the cars skid before coming to rest.

A person stands in a stationary canoe and throws a \(5.00-\mathrm{kg}\) stone with a velocity of 8.00 \(\mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. The person and canoe have a combined mass of \(105 \mathrm{~kg}\). Ignoring air resistance and effects of the water, find the horizontal recoil velocity (magnitude and direction) of the canoe.

A volleyball is spiked so that its incoming velocity of \(+4.0 \mathrm{~m} / \mathrm{s}\) is changed to an outgoing velocity of \(-21 \mathrm{~m} / \mathrm{s}\). The mass of the volleyball is \(0.35 \mathrm{~kg}\). What impulse does the player apply to the ball?

A two-stage rocket moves in space at a constant velocity of \(4900 \mathrm{~m} / \mathrm{s} .\) The two stages are then separated by a small explosive charge placed between them. Immediately after the explosion the velocity of the 1200 -kg upper stage is \(5700 \mathrm{~m} / \mathrm{s}\) in the same direction as before the explosion. What is the velocity (magnitude and direction) of the 2400 -kg lower stage after the explosion?

A baseball \((m=149 \mathrm{~g}\) ) approaches a bat horizontally at a speed of \(40.2 \mathrm{~m} / \mathrm{s}(90 \mathrm{mi} / \mathrm{h})\) and is hit straight back at a speed of \(45.6 \mathrm{~m} / \mathrm{s}(102 \mathrm{mi} / \mathrm{h})\). If the ball is in contact with the bat for a time of \(1.10 \mathrm{~ms}\), what is the average force exerted on the ball by the bat? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.
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