/*! 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 56 You push a \(2.0 \mathrm{~kg}\) ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 56

You push a \(2.0 \mathrm{~kg}\) block against a horizontal spring, compressing the spring by \(12 \mathrm{~cm}\). Then you release the block, and the spring sends it sliding across a tabletop. It stops \(75 \mathrm{~cm}\) from where you released it. The spring constant is \(170 \mathrm{~N} / \mathrm{m}\). What is the block-table coefficient of kinetic friction?

Short Answer

Expert verified
The coefficient of kinetic friction is 0.0833.

Step by step solution

01

Calculate the initial potential energy in the spring

The initial potential energy stored in the spring when it is compressed is given by the formula \[U = \frac{1}{2} k x^2\]where \(k = 170 \mathrm{~N/m}\) is the spring constant, and \(x = 0.12 \mathrm{~m}\) is the compression distance. Substituting these values in, we get:\[U = \frac{1}{2} \times 170 \times (0.12)^2 = 1.224 \mathrm{~J}\]
02

Determine the work done against friction

The work done against friction is equal to the initial potential energy stored in the spring because the spring transforms all of its potential energy to the work done on the block (neglecting other forms of energy dissipation). The block travels a distance \(d = 0.75 \mathrm{~m}\) when it stops. Thus, the work done \(W\) against friction is \[W = 1.224 \mathrm{~J}\]
03

Calculate the force of friction

The work done by friction \(W\) is given by \[W = F_{\text{friction}} \cdot d\]where \(d = 0.75 \mathrm{~m}\). Solving for \(F_{\text{friction}}\), we have:\[F_{\text{friction}} = \frac{W}{d} = \frac{1.224}{0.75} = 1.632 \mathrm{~N}\]
04

Find the coefficient of kinetic friction

The force of friction \(F_{\text{friction}}\) can be expressed in terms of the normal force \(N\) and the coefficient of kinetic friction \(\mu_k\) as:\[F_{\text{friction}} = \mu_k \cdot N\]Since the block is sliding on a horizontal surface, the normal force \(N = mg = 2.0 \times 9.8 \mathrm{~N}\). Therefore, \(N = 19.6 \mathrm{~N}\). Solving for \(\mu_k\), we have:\[\mu_k = \frac{F_{\text{friction}}}{N} = \frac{1.632}{19.6} = 0.0833\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Spring Potential Energy
The concept of spring potential energy comes into play when dealing with springs being compressed or stretched. This type of potential energy is based on Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. When you push or pull on a spring, you store energy within it. This energy is termed as potential energy because it has the potential to do work or cause motion once released.

For a spring with constant, the potential energy (PE) is calculated using the equation:
  • \[ U = \frac{1}{2} k x^2 \]
Here, \(k\) is the spring constant that measures the stiffness of the spring, and \(x\) is the displacement from the equilibrium position. The initial stored energy is completely dependent on these two values.

In the current exercise, the stored potential energy when the spring is compressed by 12 cm is 1.224 Joules. This value is the energy ready to be transferred to the block, setting it into motion once released.
Kinetic Friction
As soon as the spring releases its potential energy to push the block, kinetic friction comes into action. This force opposes the motion of the block across the table. Kinetic friction is different from static friction, as it acts between moving surfaces.

Kinetic friction depends on:
  • The type of surfaces in contact (represented by the coefficient of kinetic friction \(\mu_k\))
  • The normal force (perpendicular force) acting on the moving object
The force due to kinetic friction \(F_{\text{friction}}\) is given by the equation:
  • \[ F_{\text{friction}} = \mu_k \cdot N \]
In a horizontal motion, the normal force \(N\) is equal to the gravitational force acting on the object (mass \(m\) times the gravitational acceleration \(g\)). The resistance due to friction is what eventually brings the block to a halt after it slides 75 cm.
Work-Energy Principle
The work-energy principle is a valuable concept in physics that explains the relationship between work and energy changes in a system. According to this principle, work done on an object is equal to the change in its kinetic energy. In other words, energy transferred to an object via work will result in either an increase or a decrease of that object's kinetic energy.

In this exercise, the initial spring potential energy is completely used to overcome the kinetic friction as the block slides across the table. This means that the entire energy from the spring converts into work done against friction, eventually bringing the block to a stop.

The work-energy theorem is nicely exemplified here:
  • The work done \(W\) against friction is calculated as \(1.224 \text{ J}\), precisely the amount of initial potential energy stored in the spring.
By understanding these steps, it becomes clearer how the various energies and forces interact to determine the block's final stopping point, highlighting an applied side of the work-energy principle.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A block of mass \(6.0 \mathrm{~kg}\) is pushed up an incline to its top by a man and then allowed to slide down to the bottom. The length of incline is \(10 \mathrm{~m}\) and its height is \(5.0 \mathrm{~m}\). The coefficient of friction between block and incline is \(0.40\). Calculate (a) the work done by the gravitational force over the complete round trip of the block, (b) the work done by the man during the upward journey, (c) the mechanical energy loss due to friction over the round trip, and (d) the speed of the block when it reaches the bottom.

At a lunar base, a uniform chain hangs over the edge of a horizontal platform. A machine does \(1.0 \mathrm{~J}\) of work in pulling the rest of the chain onto the platform. The chain has a mass of \(2.0\) \(\mathrm{kg}\) and a length of \(3.0 \mathrm{~m}\). What length was initially hanging over the edge? On the Moon, the gravitational acceleration is \(1 / 6\) of \(9.8 \mathrm{~m} / \mathrm{s}^{2}\).

A \(2.0 \mathrm{~kg}\) breadbox on a frictionless incline of angle \(\theta=40^{\circ}\) is connected, by a cord that runs over a pulley, to a light spring of spring constant \(k=105 \mathrm{~N} / \mathrm{m}\), as shown in Fig. 8-31. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved \(10 \mathrm{~cm}\) down the incline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box's acceleration at the instant the box momentarily stops?

At \(t=0\) a \(1.0 \mathrm{~kg}\) ball is thrown from a tall tower with \(\vec{v}=(18 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(24 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). What is \(\Delta U\) of the ball-Earth system between \(t=0\) and \(t=6.0 \mathrm{~s}\) (still free fall)?

A child whose weight is \(267 \mathrm{~N}\) slides down a \(6.5 \mathrm{~m}\) playground slide that makes an angle of \(20^{\circ}\) with the horizontal. The coefficient of kinetic friction between slide and child is \(0.10\). (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of \(0.457 \mathrm{~m} / \mathrm{s}\), what is her speed at the bottom?
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Kinematics Physics

Read Explanation

Measurements

Read Explanation

Fundamentals of Physics

Read Explanation

Force

Read Explanation

Electrostatics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.