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Chapter 7: Problem 79

A \(250 \mathrm{~g}\) object on a frictionless, horizontal lab table is pushed against a spring of force constant \(35 \mathrm{~N} / \mathrm{cm}\) and then released. Just before the object is released, the spring is compressed \(12.0 \mathrm{~cm} .\) How fast is the object moving when it has gained half of the spring's original stored energy?

Short Answer

Expert verified
The object is moving at approximately 10.04 m/s.

Step by step solution

01

Understand the Problem

We have a mass of object \( m = 250 \text{ g} \) (which is \( 0.25 \text{ kg} \) when converted to kilograms) compressed against a spring with a force constant \( k = 35 \text{ N/cm} \) (which is \( 3500 \text{ N/m} \) when converted to meters). The spring is initially compressed by \( x = 0.12 \text{ m} \). We need to find the speed of the object after half of the spring's potential energy is converted into kinetic energy.
02

Calculate Initial Potential Energy in the Spring

The potential energy stored in a compressed or stretched spring is given by the formula: \[ U = \frac{1}{2}kx^2 \]Substitute \( k = 3500 \text{ N/m} \) and \( x = 0.12 \text{ m} \) into the equation to find the initial potential energy:\[ U = \frac{1}{2} \times 3500 \times (0.12)^2 = 25.2 \text{ J} \]
03

Calculate Half of the Spring's Original Energy

Half of the spring's original potential energy is:\[ U_{\text{half}} = \frac{25.2}{2} = 12.6 \text{ J} \]
04

Apply Conservation of Energy

According to the conservation of energy principle, the loss in potential energy will be equal to the gain in kinetic energy when the object has gained half of the spring's original stored energy:\[ \frac{1}{2}mv^2 = U_{\text{half}} \]Substitute \( m = 0.25 \text{ kg} \) and \( U_{\text{half}} = 12.6 \text{ J} \) to solve for \( v \):\[ \frac{1}{2} \times 0.25 \times v^2 = 12.6 \text{ J} \]
05

Solve for the Velocity

Rearrange the equation from Step 4 to find the velocity \( v \):\[ v^2 = \frac{2 \times 12.6}{0.25} \]\[ v^2 = 100.8 \]\[ v = \sqrt{100.8} \approx 10.04 \text{ m/s} \]

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Key Concepts

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

Spring Potential Energy
When a spring is either compressed or stretched, it stores potential energy. This type of energy is known as spring potential energy and is crucial for understanding what happens in scenarios involving springs. Imagine a spring that is squeezed tightly: it has the capacity to push back when the compression is released, thanks to stored energy.

The formula used to calculate spring potential energy is:
  • \( U = \frac{1}{2} k x^2 \)
where:
  • \( U \) is the potential energy in joules (J)
  • \( k \) is the spring constant determining the stiffness of the spring (measured in newtons per meter, N/m)
  • \( x \) is the displacement from the equilibrium position (meters, m)
To apply this in a problem, you substitute the values for \( k \) and \( x \). With a greater \( k \), the spring stores more energy for the same compression distance.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When a spring releases its stored potential energy, this energy transforms into kinetic energy as the object moves away from its position against the spring.

The formula to calculate kinetic energy is given by:
  • \( KE = \frac{1}{2} mv^2 \)
where:
  • \( KE \) is the kinetic energy in joules (J)
  • \( m \) is the mass of the object (in kilograms, kg)
  • \( v \) is the velocity of the object (in meters per second, m/s)
When an object is released from a spring, this equation helps to calculate how fast the object is moving by rearranging the formula to solve for velocity \( v \). The transition from potential energy to kinetic energy illustrates the principles of energy conservation.
Frictionless Surface
A frictionless surface implies that no energy is lost to friction as an object moves. In physics problems, this concept simplifies calculations, as you don't need to account for energy loss due to friction, allowing all potential energy to convert into kinetic energy.

In the given scenario, the object moves across a frictionless lab table. This means the calculations for potential energy turning into kinetic energy assume no loss, enabling simpler and more direct computations.

Real-world applications rarely encounter entirely frictionless surfaces, but in theoretical physics problems, they make it easier to understand the underlying principles of motion and energy conversion.
Physics Problems
Physics problems involving springs, energy, and motion are fundamental for grasping the laws governing physical systems. Breaking down these problems step-by-step, as seen in the original exercise, helps students apply formulas to real-world situations.

To solve such problems:
  • Identify what is known (mass, spring constant, compression distance) and what you need to find (e.g., speed).
  • Use relevant physics formulas (for potential and kinetic energy).
  • Take advantage of problem constraints like frictionless surfaces.
  • Apply the conservation of energy to equate potential and kinetic energy changes.
By understanding these steps and concepts, students enhance problem-solving skills, making complex physics more approachable.

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Most popular questions from this chapter

A certain spring stores \(10.0 \mathrm{~J}\) of potential energy when it is stretched by \(2.00 \mathrm{~cm}\) from its equilibrium position. (a) How much potential energy would the spring store if it were stretched an additional \(2.00 \mathrm{~cm} ?\) (b) How much potential energy would it store if it were compressed by \(2.00 \mathrm{~cm}\) from its equilibrium position? (c) How far from the equilibrium position would you have to stretch the string to store \(20.0 \mathrm{~J}\) of potential energy? (d) What is the force constant of this spring?

II A slingshot obeying Hooke's law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back \(20.0 \mathrm{~cm}\) against the elastic band, the pebble goes \(6.0 \mathrm{~m}\) high. (a) Assuming that air drag is negligible, how high will the pebble go if you pull it back \(40.0 \mathrm{~cm}\) instead? (b) How far must you pull it back so it will reach \(12.0 \mathrm{~m} ?\) (c) If you pull a pebble that is twice as heavy back \(20.0 \mathrm{~cm},\) how high will it \(\mathrm{go} ?\)

A spring with spring constant \(k\) is anchored to the wall on one side of a hockey rink. A hockey puck is pressed against the spring and then released to slide across the ice. In the process the hockey puck gains a kinetic energy \(K .\) Derive an expression for the initial compression of the spring \(x\) in terms of \(k\) and \(K\).

A \(1.50 \mathrm{~kg}\) brick is sliding along on a rough horizontal surface at \(13.0 \mathrm{~m} / \mathrm{s}\). If the brick stops in \(4.80 \mathrm{~s}\), how much mechanical energy is lost, and what happens to this energy?

A \(12.0 \mathrm{~N}\) package of whole wheat flour is suddenly placed on the pan of a scale such as you find in grocery stores. The pan is supported from below by a vertical spring of force constant \(325 \mathrm{~N} / \mathrm{m}\). If the pan has negligible weight, find the maximum distance the spring will be compressed if no energy is dissipated by friction.
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