/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 A block with mass 0.50 \(\mathrm... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 43

A block with mass 0.50 \(\mathrm{kg}\) is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 \(\mathrm{m}(\mathrm{Fig} .7 .31) .\) When released, the block moves on a horizontal tabletop for 1.00 \(\mathrm{m}\) before coming to rest. The spring constant \(k\) is 100 \(\mathrm{N} / \mathrm{m}\) . What is the coefficient of kinetic friction \(\mu_{k}\) between the block and the tabletop?

Short Answer

Expert verified
The coefficient of kinetic friction is approximately 0.408.

Step by step solution

01

Calculate the Potential Energy Stored in the Spring

The potential energy stored in a compressed spring can be calculated using the formula \( U = \frac{1}{2} k x^2 \), where \( k \) is the spring constant, and \( x \) is the compression distance. Substituting the given values: \[ U = \frac{1}{2} \times 100 \, \text{N/m} \times (0.20 \, \text{m})^2 = 2 \, \text{J}. \] This is the energy stored in the spring.
02

Determine the Work Done by Friction

The work done by friction as the block moves 1.00 m and comes to rest is equivalent to the potential energy of the spring. Thus, \[ W_{friction} = 2 \, \text{J}. \] This work is given by \( W_{friction} = f_k \cdot d \), where \( f_k \) is the frictional force and \( d \) is the distance (1.00 m). Substituting: \[ f_k \cdot 1.00 \, \text{m} = 2 \, \text{J}. \] So, \( f_k = 2 \, \text{N}. \)
03

Relate Frictional Force to Normal Force and Coefficient of Friction

The frictional force \( f_k \) is also given by the equation \( f_k = \mu_k N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force. Since the block is moving on a horizontal table, \( N = mg \), where \( m = 0.50 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). Therefore, \[ N = 0.50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 4.9 \, \text{N}. \]
04

Solve for the Coefficient of Kinetic Friction

Substituting \( f_k = 2 \, \text{N} \) and \( N = 4.9 \, \text{N} \) into the friction equation \( 2 \, \text{N} = \mu_k \times 4.9 \, \text{N} \), we solve for \( \mu_k \): \[ \mu_k = \frac{2 \, \text{N}}{4.9 \, \text{N}} \approx 0.408. \] So, the coefficient of kinetic friction is approximately \( 0.408 \).

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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 spring is compressed or stretched from its natural length, it stores energy. This energy is known as spring potential energy. Spring potential energy is a form of mechanical energy, and can be calculated using the formula \[ U = \frac{1}{2} k x^2 \]where:
  • \( U \) is the potential energy in joules (J),
  • \( k \) is the spring constant in newtons per meter (N/m),
  • \( x \) is the compression or elongation in meters (m).
For example, our exercise involved a block compressing a spring by 0.20 meters with a spring constant of 100 N/m. Substituting the values into the formula, we find the spring potential energy to be 2 J. Understanding this concept is crucial, as it shows how potential energy is stored in the spring's tension, ready to be transformed into other energy forms.
Kinetic Friction
Kinetic friction is the force that opposes the motion of two objects sliding past each other. It is crucial to understand that kinetic friction only acts once the objects are in motion, as opposed to static friction, which acts when objects are stationary. The frictional force can be calculated using:\[ f_k = \mu_k \cdot N \]where:
  • \( f_k \) is the kinetic frictional force in newtons (N),
  • \( \mu_k \) is the coefficient of kinetic friction (dimensionless),
  • \( N \) is the normal force in newtons (N), equal to the gravitational force when on a horizontal plane.
In our scenario, the coefficient of friction \( \mu_k \) determines how easily the block slides over the tabletop. Given the block's mass and gravity, the normal force is found to be 4.9 N. With the measured frictional force contributing to stopping the block, we can conclude the coefficient of kinetic friction is approximately 0.408. This tells us about the roughness or slipperiness between the block and the table's surface.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics. It states that the work done by the forces acting on an object results in a change in energy. Essentially, the principle links the work done on an object to its change in kinetic and potential energy. In a situation where a block compresses a spring, as in our exercise, the work-energy principle underpins the transition from the stored spring potential energy to overcoming kinetic friction as the block moves. When the block is released, the spring potential energy is converted into kinetic energy, propelling the block forward. As it slides across the table, kinetic friction gradually dissipates this energy, eventually bringing the block to a halt. This illustrates how the initial spring potential energy was used entirely to do the work against friction to stop the block. Thus, understanding the work-energy principle helps us comprehend how energy transformation and work done in friction are related in physical systems.

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Most popular questions from this chapter

You are designing a delivery ramp for crates containing exercise equipment. The \(1470-\mathrm{N}\) crates will move at 1.8 \(\mathrm{m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a \(550-\mathrm{N}\) kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 8.0 \(\mathrm{m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the force constant of the spring that will be needed in order to meet the design criteria.

The potential energy of a pair of hydrogen atoms separated by a large distance \(x\) is given by \(U(x)=-C_{6} / x^{6},\) where \(C_{6}\) is a positive constant. What is the force that one atom exerts on the other? Is this force attractive or repulsive?

A small rock with mass 0.20 \(\mathrm{kg}\) is released from rest at point \(A,\) which is at the top edge of a large, hemispherical bowl with radius \(R=0.50 \mathrm{m}\) (Fig. 7.25\()\) . Assume that the size of the rock is small compared to \(R,\) so that the rock can be treated as a particle, and assume that the rock slides rather than rolls. The work done by friction on the rock when it moves from point \(A\) to point \(B\) at the bottom of the bowl has magnitude 0.22 \(\mathrm{J}\) . (a) Between points \(A\) and \(B,\) how much work is done on the rock by (i) the normal force and (ii) gravity? (b) What is the speed of the rock as it reaches point \(B ?\) (c) Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? Explain. (d) Just as the rock reaches point \(B\) , what is the normal force on it due to the bottom of the bowl?

A Hooke's law force \(-k x\) and a constant conservative force \(F\) in the \(+x\) -direction act on an atomic ion. (a) Show that a possible potential-energy function for this combination of forces is \(U(x)=\frac{1}{2} k x^{2}-F x-F^{2} / 2 k\) . Is this the only possible function? Explain. (b) Find the stable equilibrium position. (c) Graph \(U(x)\) (in units of \(F^{2} / k )\) versus \(x\) (in units of \(F / k )\) for values of \(x\) between \(-5 F / k\) and 5\(F / k\) . (d) Are there any unstable equilibrium positions? (e) If the total energy is \(E=F^{2} | k,\) what are the maximum and minimum valnes of \(x\) that the ion reaches in its motion? If the ion has mass \(m,\) find its maximum speed if the total energy is \(E=F^{2} / k .\) For what value of \(x\) is the speed maximum?

A variable force \(\overrightarrow{\boldsymbol{k}}\) is maintained tangent to a frictionless semicircular surface (Fig. 7.41\() .\) By slow variations in the force, a block with weight \(w\) is moved, and the spring to which it is attached is stretched from position 1 to position \(2 .\) The spring has negligible mass and force constant \(k .\) The end of the spring moves in an are of radius \(a\) . Calculate the work done by the force \(\vec{F}\) .
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