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Chapter 7: Problem 19

A spring of negligible mass has force constant \(k=\) 1600 \(\mathrm{N} / \mathrm{m}\) . (a) How far must the spring be compressed for 3.20 \(\mathrm{J}\) of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a \(1.20-\mathrm{kg}\) book onto it from a height of 0.80 \(\mathrm{m}\) above the top of the spring. Find the maximum distance the spring will be compressed.

Short Answer

Expert verified
(a) The spring must be compressed by 0.0633 m. (b) The maximum compression of the spring is 0.1086 m.

Step by step solution

01

Understanding Spring Potential Energy Formula

The potential energy stored in a compressed or stretched spring is given by the formula \( PE = \frac{1}{2} k x^2 \), where \( PE \) is the potential energy, \( k \) is the spring constant, and \( x \) is the compression distance. In part (a), we need to find \( x \) when \( PE = 3.20 \) J.
02

Solving for Compression Distance

Rearrange the spring potential energy formula to solve for \( x \): \( x = \sqrt{\frac{2 \cdot PE}{k}} \). Substituting \( PE = 3.20 \) J and \( k = 1600 \) N/m, we get: \( x = \sqrt{\frac{2 \cdot 3.20}{1600}} \approx 0.0633 \) m. The spring must be compressed by approximately 0.0633 meters.
03

Understanding Energy Conversion for Part (b)

When the book is dropped, its gravitational potential energy is converted into spring potential energy at the maximum compression point of the spring. We equate the gravitational potential energy (GPE) with spring potential energy (SPE). The GPE is given by \( mgh \), where \( m \) is mass, \( g \) is gravity (9.81 m/s虏), and \( h \) is height.
04

Calculating Initial Gravitational Potential Energy

Compute the gravitational potential energy when the book is dropped from \( h = 0.80 \) m: \( GPE = mgh = 1.20 \times 9.81 \times 0.80 = 9.408 \) Joules.
05

Solving for Maximum Compression with Energy Conservation

Using energy conservation, set \( GPE = SPE \): \( \frac{1}{2} k x^2 = 9.408 \). Solve for \( x \), \( x = \sqrt{\frac{2 \cdot 9.408}{1600}} \approx 0.1086 \) m. The maximum compression of the spring is approximately 0.1086 meters.

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Key Concepts

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

Energy Conservation
Energy conservation is an important concept in physics that implies energy cannot be created or destroyed but can only change forms. When dealing with springs and gravity, this principle becomes useful in predicting how energy moves between objects. For example, when you compress a spring by dropping a book onto it, the gravitational potential energy (GPE) of the book transforms into the spring's potential energy (SPE).
This means the energy the book has due to its height is transferred into the spring, causing it to compress. The key is to remember that the initial energy from the book converts into the energy stored in the spring, showing perfect energy transformation from gravitational to elastic potential energy.
Spring Constant
The spring constant, symbolized as k, is a measure of a spring's stiffness. A higher spring constant indicates a stiffer spring which requires more force to compress it a certain distance.
This constant is fundamental in calculating the potential energy stored in a spring. For instance, the spring in the exercise has a constant of 1600 N/m, suggesting it鈥檚 quite stiff. In mathematical terms, this constant is used in the spring potential energy formula:
  • Potential Energy (PE) = \( \frac{1}{2} k x^2 \)
where \( x \) is the displacement from equilibrium. Thus, understanding the spring constant helps predict how much energy it can store when compressed or stretched.
Gravitational Potential Energy
Gravitational Potential Energy (GPE) depends on three factors: the mass of the object, the height from which it falls, and the force of gravity. Mathematically, we express it as:
  • GPE = \( mgh \)
where \( m \) stands for mass, \( g \) is the gravitational acceleration (approximately 9.81 m/s虏 on Earth), and \( h \) is the height.
In our exercise, when the 1.20 kg book is dropped from a height of 0.8 meters, it possesses gravitational potential energy. Once the book lands on the spring, this energy is converted into the spring's potential energy, allowing us to calculate the maximum compression of the spring using energy conservation principles. Understanding GPE helps in predicting how much spring compression will occur, given the height and mass of the falling object.

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Most popular questions from this chapter

A \(2.00-\mathrm{kg}\) block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) , compressing it 0.220 \(\mathrm{m}\) . When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope \(37.0^{\circ}\) (Fig. 7.30\()\) . (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

A spring stores potential energy \(U_{0}\) when it is compressed a distance \(x_{0}\) from its uncompressed length. (a) In terms of \(U_{0},\) how much energy does it store when it is compressed (i) twice as much and (ii) half as much? (b) In terms of \(x_{0},\) how much must it be compressed from its uncompressed length to store (i) twice as much energy and (ii) half as much energy?

On a horizontal surface, a crate with mass 50.0 \(\mathrm{kg}\) is placed against a spring that stores 360 \(\mathrm{J}\) of energy. The spring is released, and the crate slides 5.60 \(\mathrm{m}\) before coming to rest. What is the speed of the crate when it is 2.00 \(\mathrm{m}\) from its initial position?

You are designing a delivery ramp for crates containing exercise equipment. The \(1470-\mathrm{N}\) crates will move at 1.8 \(\mathrm{m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a \(550-\mathrm{N}\) kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 8.0 \(\mathrm{m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the force constant of the spring that will be needed in order to meet the design criteria.

A 10.0 -kg microwave oven is pushed 8.00 \(\mathrm{m}\) up the sloping surface of a loading ramp inclined at an angle of \(36.9^{\circ}\) above the horizontal, by a constant force \(\vec{F}\) with a magnitude 110 \(\mathrm{N}\) and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.250 . (a) What is the work done on the oven by the force \(\vec{F} ?\) (b) What is the work done on the oven by the friction force? (c) Compute the increase in potential energy for the oven. (d) Use your answers to parts (a), \((b),\) and (c) to calculate the increase in the oven's kinetic energy. \((e)\) Use \(\Sigma \overrightarrow{\boldsymbol{F}}=m \overrightarrow{\boldsymbol{a}}\) to calculate the acceleration of the oven. Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling 8.00 \(\mathrm{m}\) . From this, compute the increase in the oven's kinetic energy, and compare it to the answer you got in part (d).
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