/*! 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 39 A block with mass \(0.50 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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A block with mass \(0.50 \mathrm{~kg}\) is forced against a horizontal spring of negligible mass, compressing the spring a distance of \(0.20 \mathrm{~m}\) (Fig. \(\mathbf{P 7 . 3 9}\) ). When released, the block moves on a horizontal tabletop for \(1.00 \mathrm{~m}\) before coming to rest. The force constant \(k\) is \(100 \mathrm{~N} / \mathrm{m}\). What is the coefficient of kinetic friction \(\mu_{\mathrm{k}}\) between the block and the tabletop?

Short Answer

Expert verified
The coefficient of kinetic friction between the block and the tabletop is 0.41.

Step by step solution

01

Calculate the Spring's Potential Energy

First, calculate the potential energy stored in the spring using the formula for spring potential energy: \( E_{sp} = 0.5 * k * x^2 \), where \( k = 100 N/m \) is the spring constant and \( x = 0.20 m \) is the displacement. Substituting the values into the formula gives: \( E_{sp} = 0.5 * 100 N/m * (0.20 m)^2 = 2.0 J \). The spring potential energy is 2.0 Joules.
02

Calculate the Work Done Against Friction

As the block slides on the tabletop and comes to a stop, all its initial energy provided by the spring is used to do work against friction. Therefore, the work done against friction is equal to the spring potential energy, which is 2.0 Joules.
03

Determine the Coefficient of Kinetic Friction

Next, use the work-energy principle. The work done against friction is the product of the frictional force and the distance travelled, denoted as \( W = F_{fr} * d \). The frictional force is also the product of the normal force and the coefficient of kinetic friction, \( F_{fr} = \mu_{k} * F_{N} \), where \( F_{N} = m * g \) is the normal force, with \( m = 0.50 kg \) being the mass of the block and \( g = 9.8 m/s^2 \) the acceleration due to gravity. Substitute \( F_{fr} \) into the work equation: \( W = \mu_{k} * m * g * d \). Now, solve for \( \mu_{k} \): \( \mu_{k} = W / (m * g * d) \). Substituting the known values gives: \( \mu_{k} = 2.0 J / (0.50 kg * 9.8 m/s^2 * 1.00 m) = 0.41 \). Thus, the coefficient of kinetic friction is 0.41.

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Key Concepts

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

Spring Potential Energy
When you compress or stretch a spring, it stores energy in the form of spring potential energy. This energy depends on two key factors: the stiffness of the spring (known as the spring constant, denoted as \( k \)) and how much you compress or stretch it (displacement, denoted as \( x \)).
The energy stored in the spring is calculated using the formula:
  • \( E_{sp} = 0.5 \times k \times x^2 \)
Where:
  • \( E_{sp} \) is the spring potential energy.
  • \( k \) is the spring constant, with units Newton per meter (N/m).
  • \( x \) is the displacement of the spring from its equilibrium position.
In our exercise, we found that compressing a spring with a constant of 100 N/m by 0.20 m stores 2 Joules of energy. This stored energy is what propels the block forward once released. Understanding this is key to building a foundation in mechanics as it shows how potential energy can be converted into other forms.
Work-Energy Principle
The work-energy principle is a critical concept in physics that explains how energy is transferred or transformed. It states that the work done on an object is equal to the change in its energy.
This principle can be used to solve problems involving motion. When a force does work on an object, it is essentially transforming energy from one form to another.
In the context of our exercise, the spring potential energy of 2.0 Joules transforms into the work done against friction as the block moves and eventually stops. The formula capturing this idea is:
  • \( W = F_{fr} \times d \)
Where:
  • \( W \) is the work done, in Joules.
  • \( F_{fr} \) is the frictional force opposing motion.
  • \( d \) is the distance over which the force is applied.
Through the work-energy principle, we understand that all the initial energy provided by the spring is used to overcome friction, allowing us to solve for the coefficient of friction with clarity.
Coefficient of Kinetic Friction
The coefficient of kinetic friction, denoted as \( \mu_{k} \), specifies the degree to which friction resists the movement of two surfaces sliding past each other. It is a dimensionless value that gives us insight into the roughness between the block and the surface.
This coefficient is crucial for calculating the frictional force, which is given by:
  • \( F_{fr} = \mu_{k} \times F_{N} \)
Where:
  • \( F_{fr} \) is the frictional force.
  • \( F_{N} \) is the normal force, often equivalent to the weight of the object if on a horizontal plane.
In our exercise, by applying the work-energy principle and understanding how friction affects motion, we calculated the coefficient of kinetic friction to be 0.41. This value gives us a clear picture of how much frictional force opposed the block's motion, leading to its eventual rest.

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Most popular questions from this chapter

In one day, a 75 kg mountain climber ascends from the 1500 m level on a vertical cliff to the top at 2400 m. The next day, she descends from the top to the base of the cliff, which is at an elevation of 1350 m. What is her change in gravitational potential energy (a) on the first day and (b) on the second day?

\(A 2.50 \mathrm{~kg}\) block on a horizontal floor is attached to a horizontal spring that is initially compressed \(0.0300 \mathrm{~m}\). The spring has force constant \(840 \mathrm{~N} / \mathrm{m}\). The coefficient of kinetic friction between the floor and the block is \(\mu_{k}=0.40 .\) The block and spring are released from rest, and the block slides along the floor. What is the speed of the block when it has moved a distance of \(0.0200 \mathrm{~m}\) from its initial position? (At this point the spring is compressed \(0.0100 \mathrm{~m}\).)

A small box with mass \(0.600 \mathrm{~kg}\) is placed against a compressed spring at the bottom of an incline that slopes upward at \(37.0^{\circ}\) above the horizontal. The other end of the spring is attached to a wall. The coefficient of kinetic friction between the box and the surface of the incline is \(\mu_{k}=0.400\) The spring is released and the box travels up the incline, leaving the spring behind. What minimum elastic potential energy must be stored initially in the spring if the box is to travel \(2.00 \mathrm{~m}\) from its initial position to the top of the incline?

CALC A force parallel to the \(x\) -axis acts on a particle moving along the \(x\) -axis. This force produces potential energy \(U(x)\) given by \(U(x)=\alpha x^{4},\) where \(\alpha=0.630 \mathrm{~J} / \mathrm{m}^{4} .\) What is the force (magnitude and direction) when the particle is at \(x=-0.800 \mathrm{~m} ?\)

CALC A small block with mass \(0.0400 \mathrm{~kg}\) is moving in the \(x y\) -plane. The net force on the block is described by the potential-energy function \(U(x, y)=\left(5.80 \mathrm{~J} / \mathrm{m}^{2}\right) x^{2}-\left(3.60 \mathrm{~J} / \mathrm{m}^{3}\right) \mathrm{y}^{3}\). What are the magnitude and direction of the acceleration of the block when it is at the point \((x=0.300 \mathrm{~m}, y=0.600 \mathrm{~m}) ?\)

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