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A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant \(k=\) \(400 \mathrm{~N} / \mathrm{m} ;\) the other end of the spring is fixed in place. The cookie has a kinetic energy of \(20.0 \mathrm{~J}\) as it passes through the spring's equilibrium position. As the cookie slides, a frictional force of magnitude \(10.0 \mathrm{~N}\) acts on it. (a) How far will the cookie slide from the equilibrium position before coming momentarily to rest? (b) What will be the kinetic energy of the cookie as it slides back through the equilibrium position?

Short Answer

Expert verified
(a) The cookie slides 0.293 m before stopping. (b) The kinetic energy on return is 14.14 J.

Step by step solution

01

Understand the Problem

We have a mass attached to a spring with a known spring constant \(k = 400 \, \text{N/m}\). The mass has an initial kinetic energy of \(20.0 \, \text{J}\) at the equilibrium position, and it faces a frictional force of \(10.0 \, \text{N}\). The question asks us to find the distance it travels before coming to rest (part a) and its kinetic energy when it passes back through the equilibrium (part b).
02

Energy Conservation Equation Setup

When the cookie moves from the equilibrium position to the maximum displacement, its kinetic energy is converted to potential energy in the spring and work done against friction. At rest, total energy = potential energy in spring + work done by friction. Set up the equation: \[ KE_i = PE + W_{friction} \] where \( KE_i = 20.0 \, \text{J} \), \( PE = \frac{1}{2} k x^2 \), \( W_{friction} = f_d \cdot d \).
03

Equation for Maximum Displacement

Since we know the work done by friction is \( W_{friction} = f_d \cdot d \) (where \(f_d = 10.0 \, \text{N}\)), and \(x = d\), we incorporate friction into the energy balance: \[ 20.0 = \frac{1}{2} (400) d^2 + 10d \]. This simplifies to the quadratic equation: \[ 200d^2 + 10d = 20 \].
04

Solve Quadratic Equation for Distance

Solve the quadratic equation \[ 200d^2 + 10d - 20 = 0 \]. Divide the entire equation by 10 to simplify: \[ 20d^2 + d - 2 = 0 \]. Use the quadratic formula: \[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 20 \), \( b = 1 \), and \( c = -2 \).
05

Calculate the Displacement

Substitute the values into the quadratic formula: \[ d = \frac{-1 \pm \sqrt{1 + 160}}{40} = \frac{-1 \pm \sqrt{161}}{40} \]. Calculate \( \sqrt{161} \approx 12.6886 \), leading to solutions \(d = \frac{-1 + 12.6886}{40} \approx 0.293 \text{ m}\) (discarding the negative solution).
06

Analyze the Return Kinetic Energy

As the cookie slides back to the equilibrium position, it again converts potential energy in the spring back to kinetic energy minus work done against friction. Energy conservation as it returns: \[ KE_f = KE_i - 2W_{friction} = 20 - 10 \times 2d \]. Substitute the previous value of \(d\), \( 2d = 0.586\), \[ KE_f = 20 - 5.86 = 14.14 \, \text{J} \].

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Key Concepts

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

Spring Constant
When dealing with springs, understanding the spring constant, denoted by the symbol \( k \), is crucial. The spring constant is a measure of the stiffness of a spring. It's expressed in units of newtons per meter (N/m), indicating how much force is needed to stretch or compress the spring by one meter.
A high spring constant means the spring is stiff, requiring more force for the same amount of stretch compared to a spring with a lower constant. In our problem, the spring constant is given as \( 400 \, \text{N/m} \). This means that for every meter the spring is stretched or compressed, a force of 400 newtons is exerted by the spring.
  • The spring follows Hooke's Law, which is reflected in the equation: \( F = kx \), where \( F \) is the force exerted by the spring, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.
  • This relation helps us calculate the potential energy stored in the spring when it is displaced, expressed by the formula: \( PE = \frac{1}{2} k x^2 \).
Thus, the stiffness of the spring plays a significant role in how energy is stored and transformed in such systems.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In our scenario, the fake cookie sliding on a horizontal surface has a kinetic energy of \( 20.0 \, \text{J} \) at the equilibrium position.
Kinetic energy is calculated using the formula:
\[ KE = \frac{1}{2}mv^2 \]
where \( m \) is the mass and \( v \) is the velocity of the object.
As the cookie moves and encounters the spring, its kinetic energy is partly converted into potential energy stored in the spring, while some energy is lost due to work done against friction.
  • Initial kinetic energy at equilibrium plays a crucial role in determining how far the cookie will slide. This energy is used to overcome obstacles like friction and to compress or stretch the spring.
  • When the cookie comes to a stop, its kinetic energy becomes zero, indicating that all energy has been converted to spring potential or used in overcoming friction.
It's important to track how energy transforms between kinetic and other forms, like potential, as this helps us solve for variables such as distance and energy loss.
Frictional Force
Frictional force is an essential factor in any motion-based problem, as it represents resistance between two surfaces. Here, a frictional force of magnitude \(10.0 \, \text{N} \) acts on the sliding cookie.
This force affects how far the cookie can move after being released and contributes to the work done as the cookie moves along the surface.
  • Frictional force can be calculated using the equation \( F_{friction} = \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force.
  • In our exercise, the calculation of work done by friction (\( W_{friction} = f_d \cdot d \)) is pivotal. This work, as a form of energy loss, affects how far the cookie will travel before coming to rest.
The presence of friction is why the cookie will not return to its original kinetic energy level during its return motion, as some energy is continually lost to frictional work.

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Most popular questions from this chapter

A \(1.50 \mathrm{~kg}\) water balloon is shot straight up with an initial 1 speed of \(3.00 \mathrm{~m} / \mathrm{s}\). (a) What is the kinetic energy of the balloon just as it is launched? (b) How much work does the gravitational force do on the balloon during the balloon's full ascent? (c) What is the change in the gravitational potential energy of the balloon-Earth system during the full ascent? (d) If the gravitational potential energy is taken to be zero at the launch point, what is its value when the balloon reaches its maximum height? (e) If, instead, the gravitational potential energy is taken to be zero at the maximum height, what is its value at the launch point? (f) What is the maximum height?

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A \(3.5 \mathrm{~kg}\) block is accelerated from rest by a compressed spring of spring constant \(640 \mathrm{~N} / \mathrm{m}\). The block leaves the spring at the spring's relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction \(\mu_{k}=0.25 .\) The frictional force stops the block in distance \(D=7.8 \mathrm{~m}\). What are (a) the increase in the thermal energy of the block-floor system, (b) the maximum kinetic energy of the block, and (c) the original compression distance of the spring?

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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?

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