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A \(2.50 \mathrm{~kg}\) textbook is forced against one end of a horizontal spring of negligible mass that is fixed at the other end and has force constant \(250 \mathrm{~N} / \mathrm{m}\), compressing the spring a distance of 0.250 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.30 .\) Use the work-energy theorem to find how far the textbook moves from its initial position before it comes to rest.

Short Answer

Expert verified
After calculating all the necessary components and solving the equation presented in Step 3, we will be able to find the distance the textbook moves from its initial position before it comes to rest.

Step by step solution

01

Calculate the spring's potential energy

The textbook is compressing the spring, giving it potential energy. This potential energy can be calculated using \[PE = \frac{1}{2}kx^2\], where \(k = 250 \mathrm{~N} / \mathrm{m}\) is the spring constant and \(x = 0.250 \mathrm{~m}\) is the distance the spring is compressed. Plugging these values into the equation provides the initial potential energy.
02

Calculate the work done by kinetic friction

As the textbook slides on a horizontal tabletop, it will experience a frictional force. This friction does work against the book, taking away its kinetic energy. The work done by kinetic friction is given by \[W_f = \mu_{\mathrm{k}} * F_N * d\], where \(\mu_{\mathrm{k}} = 0.30\) is the coefficient of kinetic friction, \(F_N = mg\) is the normal force (equal to the gravitational force for this horizontal motion), and \(d\) is the distance traveled. We know the mass of the book, the acceleration due to gravity, and we're just trying to find \(d\), the distance traveled before the book stops.
03

Apply the work-energy theorem

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the initial kinetic energy of the book \(K = PE\) (because all the potential energy will be converted into kinetic energy), and when the book stops its final kinetic energy is zero. Thus, we have \(W = K_{\mathrm{initial}} - K_{\mathrm{final}}\), or \(W_f = PE - 0\). Plugging our previous results into this equation will allow us to solve for \(d\), the distance the textbook moves before it comes to rest.

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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 textbook compresses a spring, energy is stored in the spring as potential energy. The potential energy in a spring is given by the formula:
  • \[PE = \frac{1}{2}kx^2\]
  • where \(k\) is the spring constant and \(x\) is the compression distance.
For our exercise, the textbook compresses the spring by 0.250 meters. With a spring constant \(k\) of 250 N/m, we can calculate the initial spring potential energy that the textbook has stored in the spring.
Plugging in our values, we find the potential energy to be:
  • Energy = \(\frac{1}{2} \times 250 \times (0.250)^2\)
  • This calculates to an energy of 7.8125 Joules.
This energy is crucial, as it will set the textbook in motion when it is released, converting this stored potential energy into other forms of energy as it moves.
Kinetic Friction
As the textbook slides across the table, a force called kinetic friction comes into play. This force opposes the motion of the textbook, working to slow it down. Kinetic friction depends on:
  • The normal force \(F_N\), which is the force perpendicular to the surface.
  • The coefficient of kinetic friction \(\mu_k\), a measure of how much frictional force exists between the surfaces involved.
In our problem, the normal force is caused by gravity acting on the textbook and can be calculated as:
\[ F_N = mg \], where \(m\) is the mass and \(g\) is the acceleration due to gravity, approximated at 9.8 m/s². For the textbook:
  • \( F_N = 2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)
  • Giving a normal force of 24.5 N.
The work done by the kinetic friction is:
  • \[ W_f = \mu_k \times F_N \times d \]
  • with \(\mu_k = 0.30\), \( F_N = 24.5 \, \text{N} \), and \(d\) the distance traveled.
Energy Conversion
In this scenario, we witness a classic example of energy conversion. Initially, all energy is stored as spring potential energy when the spring is compressed. As the spring is released, this potential energy converts into kinetic energy, propelling the textbook. This energy conversion process is influenced by the work-energy theorem, which states:
  • The work done on an object results in a change in its kinetic energy.
The initial potential energy of the spring (7.8125 Joules) is converted to kinetic energy just as the textbook begins to move. However, as soon as the book slides on the tabletop, kinetic friction starts converting this kinetic energy into thermal energy, slowing down the textbook.
The work done by kinetic friction is calculated by:
  • \[ W_f = \mu_k \times F_N \times d \].
Since the kinetic energy becomes zero when the textbook stops, the initial potential energy entirely converts into the work done against friction, which determines the stopping distance \(d\). Understanding these energy transformations and calculations helps solve the problem using the work-energy theorem effectively.

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Most popular questions from this chapter

One end of a horizontal spring with force constant \(130.0 \mathrm{~N} / \mathrm{m}\) is attached to a vertical wall. A \(4.00 \mathrm{~kg}\) block sitting on the floor is placed against the spring. The coefficient of kinetic friction between the block and the floor is \(\mu_{\mathrm{k}}=0.400 .\) You apply a constant force \(\vec{F}\) to the block. \(\vec{F}\) has magnitude \(F=82.0 \mathrm{~N}\) and is directed toward the wall. At the instant that the spring is compressed \(80.0 \mathrm{~cm},\) what are (a) the speed of the block, and (b) the magnitude and direction of the block's acceleration?

A \(75.0 \mathrm{~kg}\) painter climbs a ladder that is \(2.75 \mathrm{~m}\) long and leans against a vertical wall. The ladder makes a \(30.0^{\circ}\) angle with the wall. (a) How much work does gravity do on the painter? (b) Does the answer to part (a) depend on whether the painter climbs at constant speed or accelerates up the ladder?

A net force along the \(x\) -axis that has \(x\) -component \(F_{x}=-12.0 \mathrm{~N}+\left(0.300 \mathrm{~N} / \mathrm{m}^{2}\right) x^{2}\) is applied to a \(5.00 \mathrm{~kg}\) object that is initially at the origin and moving in the \(-x\) -direction with a speed of \(6.00 \mathrm{~m} / \mathrm{s}\). What is the speed of the object when it reaches the point \(x=5.00 \mathrm{~m} ?\)

A force in the \(+x\) -direction with magnitude \(F(x)=18.0 \mathrm{~N}-(0.530 \mathrm{~N} / \mathrm{m}) x\) is applied to a \(6.00 \mathrm{~kg}\) box that is sitting on the horizontal, frictionless surface of a frozen lake. \(F(x)\) is the only horizontal force on the box. If the box is initially at rest at \(x=0\) what is its speed after it has traveled \(14.0 \mathrm{~m} ?\)

You are a member of an Alpine Rescue Team. You must project a box of supplies up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_{\mathrm{k}}\). Use the work-energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g, h, \mu_{k},\) and \(\alpha\)

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