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Chapter 10: Problem 30

In preparation for shooting a ball in a pinball machine, a spring \((k=675 \mathrm{N} / \mathrm{m})\) is compressed by 0.0650 \(\mathrm{m}\) relative to its unstrained length. The ball \((m=0.0585 \mathrm{kg})\) is at rest against the spring at point A. When the spring is released, the ball slides (without rolling). It leaves the spring and arrives at point \(B\) , which is 0.300 m higher than point A. Ignore friction, and find the ball's speed at point B.

Short Answer

Expert verified
The ball's speed at point B is 6.53 m/s.

Step by step solution

01

Define the Energy Principle

We will use the conservation of mechanical energy to solve this problem. The total mechanical energy at point A (where the spring is compressed) will equal the total mechanical energy at point B (where the ball has moved higher). The energy at each point includes kinetic energy (KE), potential energy due to gravity (PE), and spring potential energy (SPE). This can be expressed as: \[ KE_A + PE_A + SPE_A = KE_B + PE_B + SPE_B \]
02

Identify Energy at Point A

At point A, the ball is at rest, so its kinetic energy is zero: \[ KE_A = \frac{1}{2} m v_A^2 = 0 \] The gravitational potential energy can be set to zero as the reference point:\[ PE_A = mgh_A = 0 \]The spring potential energy at point A is given by:\[ SPE_A = \frac{1}{2} k x^2 \] Substituting the known values:\[ SPE_A = \frac{1}{2} \times 675 \times (0.0650)^2 \]
03

Calculate Spring Potential Energy at Point A

Calculate the spring potential energy using the formula from Step 2:\[ SPE_A = \frac{1}{2} \times 675 \times (0.0650)^2 = 1.426875 \, \mathrm{J} \]
04

Identify Energy at Point B

At point B, the spring is neither compressed nor stretched, so the spring potential energy is zero:\[ SPE_B = 0 \] The gravitational potential energy at this point is:\[ PE_B = mgh_B \]And the kinetic energy at point B is:\[ KE_B = \frac{1}{2} m v_B^2 \] We need to solve for the speed \(v_B\) of the ball at point B.
05

Calculate Gravitational Potential Energy at Point B

Calculate the gravitational potential energy at point B:\[ PE_B = mgh_B = 0.0585 \times 9.8 \times 0.300 = 0.172071 \mathrm{J} \]
06

Apply Energy Conservation Equation

Substitute the computed values into the energy conservation equation from Step 1:\[ 0 + 0 + 1.426875 = \frac{1}{2} \times 0.0585 \times v_B^2 + 0.172071 + 0 \]This simplifies to:\[ 1.426875 = 0.02925 v_B^2 + 0.172071 \]
07

Solve for Ball's Speed at Point B

Rearrange the equation from Step 6 to solve for \(v_B\):\[ 1.426875 - 0.172071 = 0.02925 v_B^2 \]\[ 1.254804 = 0.02925 v_B^2 \]\[ v_B^2 = \frac{1.254804}{0.02925} \]\[ v_B = \sqrt{\frac{1.254804}{0.02925}} \]Calculating gives:\[ v_B \approx 6.527 \, \mathrm{m/s} \]
08

Final Answer

The ball's speed at point B is approximately \(6.53 \mathrm{m/s}\).

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Key Concepts

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

Mechanical Energy
Mechanical energy in physics is the sum of kinetic and potential energies in a system. It reflects the ability to do work. In a closed system, like our pinball machine exercise, the total mechanical energy should remain constant, assuming no energy is lost to factors like friction or air resistance.
  • Total mechanical energy at one point equals total mechanical energy at another.
  • Can be broken down into kinetic energy, potential energy, and sometimes spring potential energy.
In our exercise, we see how mechanical energy transitions from stored energy in a compressed spring to kinetic energy of the moving ball plus its gravitational potential energy as it rises.
Kinetic Energy
Kinetic energy is the energy an object possesses due to motion. It depends on both the mass and the speed of the object. The formula to calculate kinetic energy is:\[ KE = \frac{1}{2} m v^2 \]Let's break it down:
  • The symbol \( m \) is the mass of the object.
  • The symbol \( v \) represents the velocity of the object.
At point A in our problem, the ball was at rest, meaning its kinetic energy was zero. As it moves to point B, the kinetic energy is influenced by the energy released from the spring.
Potential Energy
Potential energy is the energy stored in an object due to its position or configuration. It is a broad term that can encompass different forms of energy storage.
  • Gravitational potential energy: Energy based on height and mass.
  • Spring potential energy: Energy stored in a compressed or stretched spring.
In the exercise, potential energy becomes crucial when considering the height change from point A to B. At point B, as the ball has moved upwards, gravitational potential energy comes into play.
Spring Potential Energy
Spring potential energy is the energy stored in a spring when it is compressed or stretched from its natural equilibrium position. Calculated using Hooke's Law, the formula is:\[ SPE = \frac{1}{2} k x^2 \]Where:
  • \( k \) is the spring constant, indicating the spring's stiffness.
  • \( x \) represents the displacement from the spring's equilibrium length.
In our exercise, the spring potential energy is initially stored when the spring is compressed. This energy transforms into kinetic energy and gravitational potential energy as the ball is launched.
Gravitational Potential Energy
Gravitational potential energy depends on an object's mass, the acceleration due to gravity, and its height from a reference point. The formula used is:\[ PE = mgh \]Here's what each variable indicates:
  • \( m \) is the object's mass.
  • \( g \) stands for the acceleration due to gravity.
  • \( h \) is the height above the reference point, typically where \( PE \) is set to zero.
In our problem, at point B, the ball has moved upwards, gaining gravitational potential energy due to its elevation above point A. This shift balances out with the other forms of energy to maintain conservation of total mechanical energy.

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Most popular questions from this chapter

Multiple-Concept Example 6 presents a model for solving this problem. As far as vertical oscillations are concerned, a certain automobile can be considered to be mounted on four identical springs, each having a spring constant of \(1.30 \times 10^{5} \mathrm{N} / \mathrm{m}\) . Four identical passengers it down inside the car, and it is set into a vertical oscillation that has a period of 0.370 s. If the mass of the empty car is 1560 \(\mathrm{kg}\) , determine the mass of each passenger. Assume that the mass of the car and its passengers is distributed evenly over the springs.

A rifle fires a \(2.10 \times 10^{-2}-\mathrm{kg}\) pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by \(9.10 \times 10^{-2} \mathrm{m}\) from its unstrained length. The pellet rises to a maximum height of 6.10 \(\mathrm{m}\) above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

A spring stretches by 0.018 m when a \(2.8-\mathrm{kg}\) object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is \(f=3.0 \mathrm{Hz}\) ?

A person who weighs 670 \(\mathrm{N}\) steps onto a spring scale in the bathroom, and the spring compresses by 0.79 \(\mathrm{cm} .\) (a) What is the spring constant? \((\text { b } \text { bhat is the weight of another person who compresses the }\) spring by 0.34 \(\mathrm{cm} ?\)

A person bounces up and down on a trampoline, while always staying in contact with it. The motion is simple harmonic motion, and it takes 1.90 s to complete one cycle. The height of each bounce above the equilibrium position is 45.0 \(\mathrm{cm}\) . Determine (a) the amplitude and (b) the angular frequency of the motion. (c) What is the maximum speed attained by the person?
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