/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 104 When a 58-g tennis ball is serve... [FREE SOLUTION] | 91Ó°ÊÓ

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When a 58-g tennis ball is served, it accelerates from rest to a speed of 45 m/s. The impact with the racket gives the ball a constant acceleration over a distance of 44 cm. What is the magnitude of the net force acting on the ball?

Short Answer

Expert verified
The net force on the ball is approximately 133.47 N.

Step by step solution

01

Identify Given Information

We are given the mass of the tennis ball as 58 g, the final velocity as 45 m/s, and the distance over which the acceleration occurs as 44 cm. First, convert units where necessary: mass = 58 g = 0.058 kg, distance = 44 cm = 0.44 m.
02

Apply Kinematic Equation

Use the kinematic equation: \[ v^2 = u^2 + 2a s \]where \( v = 45 \text{ m/s} \) is the final velocity, \( u = 0 \text{ m/s} \) is the initial velocity, \( a \) is the acceleration, and \( s = 0.44 \text{ m} \) is the distance. Solving for \( a \):\[ 45^2 = 0 + 2a(0.44) \]
03

Solve for Acceleration

Rearrange the equation from Step 2 to find acceleration:\[ a = \frac{45^2}{2 \times 0.44} \]Calculate \( a \):\[ a = \frac{2025}{0.88} \approx 2301.14 \text{ m/s}^2 \]
04

Use Newton's Second Law

Apply Newton's second law, \( F = ma \), to find the net force acting on the ball. Here \( m = 0.058 \text{ kg} \) and \( a = 2301.14 \text{ m/s}^2 \):\[ F = 0.058 \times 2301.14 \]
05

Calculate Net Force

Compute the force using the values from Step 4:\[ F = 0.058 \times 2301.14 = 133.467 \text{ N} \]

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Key Concepts

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

Kinematics
Kinematics is a branch of classical mechanics that deals with the motion of objects without considering the forces that cause the motion. It describes how the position of an object changes with time. In our exercise, the tennis ball is initially at rest and then moves with a speed of 45 m/s once it is struck. These changes in the ball's motion are described using kinematic equations.

The kinematic equation used here is \[ v^2 = u^2 + 2as \]where:
  • \( v \) is the final velocity of the ball.
  • \( u \) is the initial velocity.
  • \( a \) is the acceleration.
  • \( s \) is the distance over which the acceleration occurs.
This equation allows us to calculate the acceleration when velocity and distance are known. Understanding this equation helps in determining how forces affect motion, setting a foundation for solving problems involving motion in one dimension.
Acceleration
Acceleration is the rate of change of velocity with respect to time. In this context, it describes how quickly the tennis ball speeds up after being hit by a racket. Calculating acceleration involves finding how much the velocity of the ball increases over a specific distance.

Using the kinematic equation, we rearrange it to solve for acceleration: \[ a = \frac{v^2}{2s} \].
This means acceleration depends directly on the square of the velocity and inversely on the distance over which the acceleration takes place. In our exercise, this calculation helps us understand how fast the ball speeds up during the short time it's in contact with the racket.

The determined acceleration of approximately \( 2301.14 ext{ m/s}^2 \) highlights the rapid change in motion the tennis ball experiences with each serve.
Force calculation
Force calculation is a fundamental aspect of dynamics, and it's pivotal to understand how forces cause changes in motion. Newton's Second Law of Motion is used here, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, \[ F = ma \].

In our problem:
  • The mass \( m \) is \( 0.058 ext{ kg} \).
  • The acceleration \( a \) is \( 2301.14 ext{ m/s}^2 \).
By substituting these values into the equation, we calculate the net force acting on the tennis ball. This force calculation is critical to quantify how much of an impact the racket has when hitting the ball, which can influence the ball's speed and direction.
Mass and weight conversion
In many physics problems, the conversion of units plays an integral role. Mass and weight conversion often involve understanding how to switch between units of grams to kilograms, or centimeters to meters, to ensure consistency in calculations.

In our problem, knowing that:
  • 1 kilogram equals 1000 grams
  • 1 meter equals 100 centimeters
helps in converting the mass of the tennis ball from 58 grams to 0.058 kilograms and the distance from 44 centimeters to 0.44 meters.

Correct conversions are crucial as physics equations use standard SI units. Accurate unit conversion avoids errors and ensures the precision of your calculations.

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Most popular questions from this chapter

At a time when mining asteroids has become feasible, astronauts have connected a line between their 3500-kg space tug and a 6200-kg asteroid. Using their tug’s engine, they pull on the asteroid with a force of 490 N. Initially the tug and the asteroid are at rest, 450 m apart. How much time does it take for the tug and the asteroid to meet?

A fisherman is fishing from a bridge and is using a "45-N test line." In other words, the line will sustain a maximum force of \(45 \mathrm{N}\) without breaking. What is the weight of the heaviest fish that can be pulled up vertically when the line is reeled in (a) at a constant speed and (b) with an acceleration whose magnitude is \(2.0 \mathrm{m} / \mathrm{s}^{2} ?\)

A rescue helicopter is lifting a man (weight = 822 N) from a capsized boat by means of a cable and harness. (a) What is the tension in the cable when the man is given an initial upward acceleration of \(1.10 \mathrm{m} / \mathrm{s}^{2} ?\) (b) What is the tension during the remainder of the rescue when he is pulled upward at a constant velocity?

A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the earth and the moon. The distance between the earth and the moon is \(3.85 \times 10^{8} \mathrm{m},\) and the mass of the earth is 81.4 times as great as that of the moon.

A \(5.0-\mathrm{kg}\) rock and a \(3.0 \times 10^{-4}-\mathrm{kg}\) pebble are held near the surface of the earth. (a) Determine the magnitude of the gravitational force exerted on each by the earth. (b) Calculate the magnitude of the acceleration of each object when released.

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