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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 -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 0.1 m. (b) The spring will be compressed by approximately 0.108 m.

Step by step solution

01

Understand the Potential Energy Formula

The potential energy stored in a spring is given by the formula \( U = \frac{1}{2}kx^2 \), where \( U \) is the potential energy, \( k \) is the spring constant, and \( x \) is the compression distance. For part (a), we are given \( U = 3.20 \text{ J} \) and \( k = 1600 \text{ N/m} \). We need to solve for \( x \).
02

Solve for Compression Distance in Part (a)

Rearrange the formula \( U = \frac{1}{2}kx^2 \) to solve for \( x \): \( x = \sqrt{\frac{2U}{k}} \). Substitute the known values: \( x = \sqrt{\frac{2 \times 3.20}{1600}} \). Calculate \( x \) to get \( x = 0.1 \text{ m} \).
03

Understand the Energy Conservation for Part (b)

When the book is dropped onto the spring, kinetic energy due to the drop height is transformed into potential energy in the spring. The gravitational potential energy at the height is given by \( mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height. This will equal the spring's potential energy \( \frac{1}{2}kx^2 \) at maximum compression.
04

Solve for Maximum Compression in Part (b)

Set \( mgh = \frac{1}{2}kx^2 \) to solve for \( x \). Substitute \( m = 1.20 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), \( h = 0.80 \text{ m} \), and \( k = 1600 \text{ N/m} \). Calculate \( mgh = 1.20 \times 9.81 \times 0.80 = 9.408 \text{ J} \). Rearrange to \( x = \sqrt{\frac{2mgh}{k}} \). Substitute in the known values to find \( x = \sqrt{\frac{2 \times 9.408}{1600}} \). Calculate \( x \) to get \( x \approx 0.108 \text{ m} \).

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Key Concepts

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

Spring Potential Energy
When we talk about springs, one fundamental concept is the potential energy they can store. This energy is known as spring potential energy and is crucial for understanding how systems involving springs behave. The potential energy stored in a spring depends on two main factors: the spring constant \( k \) and the compression or extension distance \( x \).
The formula to calculate spring potential energy is given as:\[U = \frac{1}{2} k x^2\]
  • \( U \) is the potential energy stored in the spring.
  • \( k \) represents the spring constant, a measure of how stiff the spring is.
  • \( x \) is the compression or extension of the spring from its natural length.
This relationship shows us that the energy stored increases with the square of the compression or extension distance. In simpler terms, doubling \( x \) quadruples the energy stored.
Conservation of Energy
One of the most powerful principles in physics is the conservation of energy. It states that the total energy in a closed system remains constant; energy can neither be created nor destroyed but can shift forms. In our spring problem, conservation of energy is key to solving both parts. In part (a) of the solution, we are given the spring potential energy and need to deduce the compression distance. This reflects direct usage of potential energy concepts.
However, in part (b), energy conversion from gravitational energy to spring potential energy is at play.
  • Gravitational potential energy is given by \( mgh \), indicating the energy due to an object's height.
  • At maximum compression, all gravitational potential energy becomes stored in the spring as spring potential energy \( \frac{1}{2} k x^2 \).
This conversion highlights how the total energy is conserved even as it transitions from one form to another, thereby allowing us to use simple equations to solve for complex real-world problems.
Kinetic Energy Transformation
Kinetic energy transformation is another crucial aspect when examining spring-related problems. When objects fall and collide with springs, their kinetic energy is another variable that adds complexity to how we study these dynamics.Initially, as the book falls onto the spring, it primarily possesses gravitational potential energy due to its elevated position, described as \( mgh \). On hitting the spring:
  • This gravitational energy initiates as kinetic energy while in motion.
  • Upon impact, kinetic energy transforms entirely into spring potential energy at maximum compression.
The essential idea is that as an object's kinetic energy changes during motion, it impacts how energy is stored or released in springs. This energy transformation governs how far the spring compresses when loaded, aiding in resolving part (b). Understanding these transformations provides a clear insight into the behavior of systems where potential energy and kinetic energy interplay.

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Most popular questions from this chapter

A slingshot will shoot a 10 -g pebble 22.0 \(\mathrm{m}\) straight up. (a) How much potential energy is stored in the slingshot's rubberband? (b) With the same potential energy stored in the rubber band, how high can the slingshot shoot a \(25-\) g pebble? (c) What physical effects did you ignore in solving this problem?

Two blocks with different masses are attached to either end of a light rope that passes over a light, frictionless pulley suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended \(1.20 \mathrm{m},\) its speed is 3.00 \(\mathrm{m} / \mathrm{s} .\) If the total mass of the two blocks is 15.0 \(\mathrm{kg},\) what is the mass of each block?

A baseball is thrown from the roof of a 22.0 - -tall building with an initial velocity of magnitude 12.0 \(\mathrm{m} / \mathrm{s}\) and directed at an angle of \(53.1^{\circ}\) above the horizontal. (a) What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance. (b) What is the answer for part (a) if the initial velocity is at an angle of \(53.1^{\circ}\) below the horizontal? \((\mathrm{c})\) If the effects of air resistance are are included, will part (a) or (b) give the higher speed?

Up and Down the Hill. \(\wedge 28\) -kg rock approaches the foot of a hill with a speed of 15 \(\mathrm{m} / \mathrm{s}\) . This hill slopes upward at a constant angle of \(40.0^{\circ}\) above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20 , respectively. (a) Use energy conservation to find the maximum height above the foot of the hill reached by the rock. (b) Will the rock remain at rest at its highest point, or will it slide back down the hill? (c) If the rock does slide back down, find its speed when it returns to the bottom of the hill.

CALC A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x)=\alpha / x^{2}-\beta / x,\) where \(\alpha\) and \(\beta\) are positive constants. The proton is released from rest at \(x_{0}=\alpha / \beta .\) (a) Show that \(U(x)\) can be written as $$U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right]$$ Graph \(U(x) .\) Calculate \(U\left(x_{0}\right)\) and thereby locate the point \(x_{0}\) on the graph. (b) Calculate \(v(x)\) , the speed of the proton as a function of position. Graph \(v(x)\) and give a qualitative description of the motion. (c) For what value of \(x\) is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the proton be released instead at \(x_{1}=3 \alpha / \beta\) . Locate the point \(x_{1}\) on the graph of \(U(x) .\) Calculate \(v(x)\) and give a qualitative description of the motion. (f) For each release point \(\left(x=x_{0}\) and \(x=x_{1}\right),\) what are the maximum and minimum values of \(x\) reached during the motion?
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