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Chapter 10: Problem 1

What is the spring constant of a spring that stores \(25 \mathrm{~J}\) of elastic potential energy when compressed by \(7.5 \mathrm{~cm}\) from its relaxed length?

Short Answer

Expert verified
The spring constant is \(8888.89 \mathrm{~N/m}\).

Step by step solution

01

Identify Given Values

Given:- Elastic potential energy, \(E_p = 25 \mathrm{~J}\)- Compression distance, \(x = 7.5 \mathrm{~cm} = 0.075 \mathrm{~m}\)
02

Write the Elastic Potential Energy Formula

The formula for elastic potential energy stored in a spring is \[E_p = \frac{1}{2} k x^2\].
03

Rearrange the Formula to Solve for the Spring Constant

Solve for \(k\):\[E_p = \frac{1}{2} k x^2\]Multiply both sides by 2:\[2E_p = k x^2\]Divide both sides by \(x^2\):\[k = \frac{2E_p}{x^2}\]
04

Substitute the Given Values into the Formula

Substitute \(E_p = 25 \mathrm{~J}\) and \(x = 0.075 \mathrm{~m} \) into the formula for \(k\):\[k = \frac{2 \times 25}{(0.075)^2}\]
05

Calculate the Spring Constant

Calculate \(k\):\[k = \frac{50}{0.005625} = 8888.89 \mathrm{~N/m}\]

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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 in stretchy or compressible materials like springs when they are deformed. Imagine you are squeezing or stretching a spring - the more you do this, the more energy you store in the spring. This stored energy can be released to do work.

The formula for calculating the elastic potential energy stored in a spring is:

eqn. update

Here, the energy ( Ep ) depends on the spring constant ( k ) and the compression or extension distance ( x ). The spring constant tells us how stiff the spring is. A larger k value means the spring is harder to compress or extend.
Hooke's Law
Hooke's Law describes the behavior of springs. It states that the force needed to compress or extend a spring is directly proportional to the distance it is stretched or compressed. This is written as:

F = kx

Here, F is the force applied, k is the spring constant, and x is the distance the spring is stretched or compressed.

Using Hooke's Law, we understand why springs resist deformation: the greater the deformation, the larger the force needed. This principle is vital in various applications, from car suspensions to measuring instruments.
Compression Distance
Compression distance refers to how much a spring is compressed from its natural, relaxed length. In the given problem, the spring is compressed by 7.5 cm, which we convert to meters ( 0.075 m).

The compression distance ( x ) is crucial in calculating both the elastic potential energy and determining the spring constant using the formulas we've discussed.

Here鈥檚 a step-by-step breakdown on calculating the spring constant ( k ) using the provided information:

  • Identify the given values: Ep = 25 J and x = 0.075 m.
  • Use the equation for elastic potential energy: Ep = (1/2)kx^2 .
  • Rearrange the formula to solve for the spring constant ( k ):
    k = 2Ep/x^2

  • Substitute the values and calculate: k = (2 * 25)/(0.075^2) = 8888.89 N/m.

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Most popular questions from this chapter

Nonconforming Spring A certain spring is found \(n o t\) to conform to Hooke's law. The force (in newtons) it exerts when stretched a distance \(x\) (in meters) is found to have magnitude \((52.8 \mathrm{~N} / \mathrm{m}) x+\left(38.4 \mathrm{~N} / \mathrm{m}^{2}\right) x^{2}\) in the direction opposing the stretch. (a) Compute the work required to stretch the spring from \(x_{1}=0.500\) \(\mathrm{m}\) to \(x_{2}=1.00 \mathrm{~m} .\) (b) With one end of the spring fixed, a particle of mass \(2.17 \mathrm{~kg}\) is attached to the other end of the spring when it is extended by an amount \(x_{2}=1.00 \mathrm{~m}\). If the particle is then released from rest, what is its speed at the instant the spring has returned to the configuration in which the extension is \(x_{1}=0.500 \mathrm{~m} ?(\mathrm{c})\) Is the force exerted by the spring conservative or nonconservative? Explain.

Diatomic Molecule The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$ U=\frac{A}{r^{12}}-\frac{B}{r^{6}} $$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation-that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

Two Blocks and a Spring A block of mass \(m_{A}=2.0 \mathrm{~kg}\) slides along a frictionless table with a speed of \(10 \mathrm{~m} / \mathrm{s}\). Directly in front of it, and moving in the same direction, is a block of mass \(m_{B}=5.0 \mathrm{~kg}\) moving at \(3.0 \mathrm{~m} / \mathrm{s}\). A massless spring with spring constant \(k=1120\) \(\mathrm{N} / \mathrm{m}\) is attached to the near side of \(m_{B}\), as shown in Fig. \(10-57\). When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic at this point.)

Bullet Hits Wall A \(30 \mathrm{~g}\) bullet, with a horizontal velocity of \(500 \mathrm{~m} / \mathrm{s}\), comes to a stop \(12 \mathrm{~cm}\) within a solid wall. (a) What is the change in its mechanical energy? (b) What is the magnitude of the average force from the wall stopping it?

Potential Energy Function A single conservative force \(F(x)\) acts on a \(1.0 \mathrm{~kg}\) particle that moves along an \(x\) axis. The potential energy \(U(x)\) associated with \(F(x)\) is given by $$ U(x)=(-4.00 \mathrm{~J} / \mathrm{m}) e^{(-x /(4.00 \mathrm{~m}))} $$ At \(x=5.0 \mathrm{~m}\) the particle has a kinetic energy of \(2.0 \mathrm{~J}\). (a) What is the mechanical energy of the system? (b) Make a plot of \(U(x)\) as a function of \(x\) for \(0 \leq x \leq 10 \mathrm{~m}\), and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine (c) the least value of \(x\) and (d) the greatest value of \(x\) between which the particle can move. Use part (b) to determine (e) the maximum kinetic energy of the particle and (f) the value of \(x\) at which it occurs. (g) Determine the equation for \(F(x)\) as a function of \(x\). (h) For what (finite) value of \(x\) does \(F(x)=0 ?\)
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