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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 \(\mathrm{A} .\) When the spring is released, the ball slides (without rolling). It leaves the spring and arrives at point \(\mathrm{B},\) which is \(0.300 \mathrm{m}\) higher than point \(\mathrm{A} .\) Ignore friction, and find the ball's speed at point \(\mathrm{B}\).

Short Answer

Expert verified
The ball's speed at point B is approximately 6.54 m/s.

Step by step solution

01

Define Known Values

First, let's list the known values: the spring constant \(k = 675 \, \mathrm{N/m}\), the amount of compression \(x = 0.065 \, \mathrm{m}\), the mass of the ball \(m = 0.0585 \, \mathrm{kg}\), and the height difference between points A and B \(h = 0.300 \, \mathrm{m}\). These values will be used to calculate the speed of the ball at point B.
02

Calculate the Potential Energy in the Spring

The potential energy stored in the compressed spring can be found using the formula for elastic potential energy: \[ PE_{spring} = \frac{1}{2} k x^2 \]Substitute the values into the formula: \[ PE_{spring} = \frac{1}{2} \times 675 \, \mathrm{N/m} \times (0.065 \, \mathrm{m})^2 \]Calculate this to find \( PE_{spring} = 1.4259375 \, \mathrm{J}\).
03

Use Conservation of Energy Principle

Initially, the total mechanical energy at point A is only the spring potential energy. At point B, the mechanical energy is the sum of kinetic energy and gravitational potential energy. Apply the conservation of mechanical energy:\[ PE_{spring} = KE_B + PE_{gravity, B} \]The gravitational potential energy at point B is \[ PE_{gravity, B} = mgh \]Substitute known values: \[ PE_{gravity, B} = 0.0585 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s}^2 \times 0.300 \, \mathrm{m} \approx 0.172215 \, \mathrm{J} \]
04

Calculate the Kinetic Energy at Point B

Re-arrange the energy conservation equation to solve for the kinetic energy at point B:\[ KE_B = PE_{spring} - PE_{gravity, B} \]Substitute the known values into the equation:\[ KE_B = 1.4259375 \, \mathrm{J} - 0.172215 \, \mathrm{J} \approx 1.2537225 \, \mathrm{J} \]
05

Calculate Ball's Speed at Point B

Now, use the kinetic energy formula to find the speed at point B:\[ KE_B = \frac{1}{2} mv^2 \]Solve for \( v \):\[ v = \sqrt{\frac{2 \times KE_B}{m}} \]Substitute the known values:\[ v = \sqrt{\frac{2 \times 1.2537225 \, \mathrm{J}}{0.0585 \, \mathrm{kg}}} \]Calculate to find \( v \approx 6.54 \, \mathrm{m/s}\).

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Key Concepts

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

Elastic Potential Energy
Elastic potential energy is the energy stored when objects are compressed or stretched. In our example, a spring is compressed, storing potential energy that can later be converted into kinetic energy.
The formula to calculate the elastic potential energy in a spring is given by:
  • \( PE_{spring} = \frac{1}{2} k x^2 \)
where:
  • \( k \) is the spring constant, which measures the stiffness of the spring.
  • \( x \) is the displacement from the spring's original (unstressed) position.
In the pinball machine scenario, the spring constant is 675 N/m, and the compression is 0.065 m.
Using these values, we can calculate the elastic potential energy stored when the spring is compressed. This energy is then transferred to the ball, setting it into motion. Understanding how elastic potential energy converts into other forms is crucial in applications like this.
Kinetic Energy
Kinetic energy is the energy of motion. Any object that is moving has kinetic energy, which depends on its mass and speed.
To calculate kinetic energy, we use the formula:
  • \( KE = \frac{1}{2} mv^2 \)
where:
  • \( m \) is the mass of the object.
  • \( v \) is its velocity.
In this exercise, once the ball is released from the spring's compression, it gains kinetic energy. Initially, at rest, the ball's speed increases as it moves, converting the stored elastic potential energy into kinetic energy.
By using the conservation of energy principle, the calculation involves finding how much of the stored energy converts into kinetic energy after accounting for other energy forms, like gravitational potential energy. At point B, we find the ball's kinetic energy, which allows us to determine its speed as it moves to a new position.
Gravitational Potential Energy
Gravitational potential energy is the energy stored in an object as a result of its vertical position or height. This concept comes into play when an object is lifted against the force of gravity.
The formula for gravitational potential energy is:
  • \( PE_{gravity} = mgh \)
where:
  • \( m \) is the mass of the object.
  • \( g \) is the acceleration due to gravity (approximately 9.81 \( \mathrm{m/s}^2 \) on Earth).
  • \( h \) is the height difference from a reference point.
In the problem, as the ball reaches point B, it rises to a height of 0.300 m.
The gravitational potential energy at this point must be accounted for because some of the elastic potential energy initially stored in the spring is transferred to gravitational potential energy as the ball is lifted.
Understanding how gravitational potential energy affects energy conservation ensures an accurate description of the ball's speed at different stages of its motion.

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Most popular questions from this chapter

A spring stretches by \(0.018 \mathrm{m}\) 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 horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connected to a movable object. The spring and object are compressed by \(0.065 \mathrm{m},\) released from rest, and subsequently oscillate back and forth with an angular frequency of \(11.3 \mathrm{rad} / \mathrm{s} .\) What is the speed of the object at the instant when the spring is stretched by \(0.048 \mathrm{m}\) relative to its unstrained length?

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to \(3.0 \times 10^{-2} \mathrm{N}\). The length and radius of the collagen are, respectively, 2.5 and \(0.091 \mathrm{cm},\) and Young's modulus is \(3.1 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}\). (a) If the stretching obeys Hooke's law, what is the spring constant \(k\) for collagen? (b) How much work is done by the variable force that stretches the collagen? (See Section 6.9 for a discussion of the work done by a variable force.)

A vertical spring with a spring constant of \(450 \mathrm{N} / \mathrm{m}\) is mounted on the floor. From directly above the spring, which is unstrained, a \(0.30-\mathrm{kg}\) block is dropped from rest. It collides with and sticks to the spring, which is compressed by \(2.5 \mathrm{cm}\) in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in \(\mathrm{cm}\) ) above the compressed spring was the block dropped?

In a room that is \(2.44 \mathrm{m}\) high, a spring (unstrained length \(=0.30 \mathrm{m}\) ) hangs from the ceiling. A board whose length is \(1.98 \mathrm{m}\) is attached to the free end of the spring. The board hangs straight down, so that its \(1.98-\mathrm{m}\) length is perpendicular to the floor. The weight of the board \((104 \mathrm{N})\) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?
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