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Chapter 6: Problem 75

A 2.50-kg textbook is forced against a horizontal spring of negligible mass and force constant 250 N/m, compressing the spring a distance of 0.250 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \(\mu_k\) \(=\) 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
The textbook moves approximately 1.062 meters before coming to rest.

Step by step solution

01

Identify Given Values

First, let's identify all the values given in the problem.- Mass of the textbook, \( m = 2.50 \) kg- Spring constant, \( k = 250 \) N/m- Compression of the spring, \( x = 0.250 \) m- Coefficient of kinetic friction, \( \mu_k = 0.30 \)
02

Calculate Initial Spring Potential Energy

The potential energy stored in the spring when compressed is given by the formula:\[ PE_{spring} = \frac{1}{2} k x^2 \]Substitute the given values:\[ PE_{spring} = \frac{1}{2} \times 250 \times (0.250)^2 = 7.8125 \text{ J} \]
03

Determine Work Done by Friction

The work done by friction as the textbook slides can be calculated using:\[ W_{friction} = -f_k \times d \]where the frictional force \( f_k \) is \( f_k = \mu_k \times m \times g \). Here, \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity.
04

Calculate the Frictional Force

Calculate \( f_k \):\[ f_k = 0.30 \times 2.50 \times 9.81 = 7.3575 \text{ N} \]
05

Apply the Work-Energy Theorem

According to the work-energy theorem, the initial mechanical energy is equal to the work done by friction:\[ PE_{spring} = -W_{friction} \]Therefore, \[ 7.8125 = 7.3575 \times d \]
06

Solve for Distance

Now solve for \( d \):\[ d = \frac{7.8125}{7.3575} \approx 1.062 \text{ m} \]

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Key Concepts

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

Spring Potential Energy
Spring potential energy is the energy stored in a spring when it is compressed or stretched. This type of energy is a form of potential energy, meaning it has the potential to change into another form, like kinetic energy, given the right circumstances. The formula for calculating spring potential energy is
  • \( PE_{spring} = \frac{1}{2} k x^2 \)
where:
  • \( k \) is the spring constant, a measure of the spring's stiffness.
  • \( x \) is the displacement from the spring's equilibrium position.
For the exercise in question, you calculated the spring potential energy to be \( 7.8125 \) Joules using these given values. When the textbook compresses the spring 0.250 m, this energy is stored, and when released, it propels the textbook forward.
Keep this formula handy as it is essential for problems involving spring mechanics and the transfer of energy.
Kinetic Friction
Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. Unlike static friction, which acts on stationary objects, kinetic friction is experienced by moving objects.
The force of kinetic friction \( f_k \) can be calculated using the equation:
  • \( f_k = \mu_k \times m \times g \)
where:
  • \( \mu_k \) is the coefficient of kinetic friction, which is 0.30 in the provided exercise.
  • \( m \) is the mass of the object, 2.50 kg for the textbook.
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \).
In the problem, the calculated frictional force is found to be \( 7.3575 \text{ N} \). This force does negative work as the book slides, reducing the mechanical energy until it comes to rest.
Understanding kinetic friction is crucial for predicting how objects decelerate and stop, especially on various surfaces.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energy in a system. It is a measure of the ability to do work, and it's conserved in the absence of non-conservative forces like friction. In this exercise, the spring potential energy is converted into kinetic energy of the sliding textbook.
However, due to kinetic friction, some of this mechanical energy is lost as heat and sound. The work-energy theorem helps us understand this conversion by asserting that:
  • \( PE_{spring} = -W_{friction} \)
The work done by friction corresponds directly to the loss of mechanical energy, resulting in the textbook coming to a stop. By applying this theorem, you were able to calculate the distance the textbook traveled before it came to a rest: approximately \( 1.062 \text{ m} \).
Grasping the interplay between different forms of energy allows us to solve complex problems about motion and forces effectively.

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Most popular questions from this chapter

A soccer ball with mass 0.420 kg is initially moving with speed 2.00 m/s. A soccer player kicks the ball, exerting a constant force of magnitude 40.0 N in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 m/s?

A proton with mass 1.67 \(\times\) 10\(^{-27}\) kg is propelled at an initial speed of 3.00 \(\times\) 10\(^5\) m/s directly toward a uranium nucleus 5.00 m away. The proton is repelled by the uranium nucleus with a force of magnitude \(F = \alpha/x^2\), where \(x\) is the separation between the two objects and \(\alpha = 2.12 \times 10^{-26} \, \mathrm{N} \cdot \mathrm{m}^2\). Assume that the uranium nucleus remains at rest. (a) What is the speed of the proton when it is \(8.00 \times 10^{-10}\) m from the uranium nucleus? (b) As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the proton get? (c) What is the speed of the proton when it is again 5.00 m away from the uranium nucleus?

A 1.50-kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s, and at point \(B\) it has slowed to 1.25 m/s. (a) How much work was done on the book between \(A\) and \(B\)? (b) If \(-\)0.750 J of work is done on the book from \(B\) to \(C\), how fast is it moving at point \(C\)? (c) How fast would it be moving at \(C\) if \(+\)0.750 J of work was done on it from \(B\) to \(C\)?

While hovering, a typical flying insect applies an average force equal to twice its weight during each downward stroke. Take the mass of the insect to be 10 g, and assume the wings move an average downward distance of 1.0 cm during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.

(a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant k, what is the force constant of each half, in terms of \(k\)? (\(Hint\): Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal?) (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k\)?
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