/*! 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 26 A 2.50 -kg block on a horizontal... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 26

A 2.50 -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 \(\mu_{\mathrm{k}}=0.40 .\) 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 com- pressed 0.0100 \(\mathrm{m.}\) .

Short Answer

Expert verified
The speed of the block is 0.38 m/s.

Step by step solution

01

Calculate Initial Potential Energy

The potential energy stored in the spring when it is compressed is given by the formula for elastic potential energy: \[ PE_{initial} = \frac{1}{2} k x_0^2 \] where \( k = 840 \; \text{N/m} \) is the spring constant and \( x_0 = 0.0300 \; \text{m} \) is the initial compression. Calculating this gives: \[ PE_{initial} = \frac{1}{2} \times 840 \times (0.0300)^2 = 0.378 \; \text{J} \]
02

Calculate Final Potential Energy

When the block has moved 0.0200 m, the spring is still compressed by 0.0100 m. The potential energy at this point is: \[ PE_{final} = \frac{1}{2} k x^2 \] where \( x = 0.0100 \; \text{m} \). Calculating this gives: \[ PE_{final} = \frac{1}{2} \times 840 \times (0.0100)^2 = 0.042 \; \text{J} \]
03

Calculate Work Done by Friction

The work done by friction is calculated using: \[ W_{friction} = -f_k d \] where the friction force \( f_k = \mu_k mg \). Here, \( \mu_k = 0.40 \), \( m = 2.50 \; \text{kg} \), and \( g = 9.8 \; \text{m/s}^2 \). So, \( f_k = 0.40 \times 2.50 \times 9.8 = 9.8 \; \text{N} \). The distance \( d = 0.0200 \; \text{m} \). Hence, \[ W_{friction} = -9.8 \times 0.0200 = -0.196 \; \text{J} \]
04

Apply Conservation of Energy

Using the conservation of energy principle, the initial potential energy plus work done by non-conservative forces (friction) equals the final kinetic plus potential energy. That is: \[ PE_{initial} + W_{friction} = KE_{final} + PE_{final} \] We have \( 0.378 \; \text{J} - 0.196 \; \text{J} = KE_{final} + 0.042 \; \text{J} \). Solving for \( KE_{final} \) gives: \[ KE_{final} = 0.180 \; \text{J} \]
05

Calculate Final Speed

The kinetic energy is related to the speed by: \[ KE = \frac{1}{2} m v^2 \] Thus, solving for \( v \): \[ 0.180 = \frac{1}{2} \times 2.50 \times v^2 \] Rearranging gives: \[ v^2 = \frac{0.180 \times 2}{2.50} \] \[ v^2 = 0.144 \]\[ v = \sqrt{0.144} = 0.38 \; \text{m/s} \]

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Key Concepts

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

Elastic Potential Energy
When dealing with springs, especially in physics problems, the concept of elastic potential energy is crucial. Elastic potential energy refers to the energy stored in a spring when it is compressed or stretched. The energy is "potential" because it's stored and can be converted into kinetic energy or work when the spring returns to its original shape.

The formula for calculating elastic potential energy is as follows:\[ PE = \frac{1}{2} k x^2 \]
  • Here, \( k \) is the spring constant, a measure of the spring's stiffness. High \( k \) values mean a stiffer spring.
  • \( x \) is the displacement from the spring's equilibrium position, which could either be the amount it is compressed or stretched.
In our exercise, the block is attached to a spring with a stiffness of 840 N/m. Initially, the spring is compressed by 0.0300 m, giving it an initial potential energy. This energy helps the block to start moving once released.
Kinetic Friction
Kinetic friction plays a significant role when objects slide across surfaces. It's the force that resists the motion of an object sliding over a surface.

The formula for the force of kinetic friction \( f_k \) is:\[ f_k = \mu_k mg \]
  • \( \mu_k \) is the coefficient of kinetic friction, which varies depending on the materials of the contacting surfaces. In this case, it is 0.40.
  • \( m \) represents the mass of the object, here 2.50 kg.
  • \( g \) is the acceleration due to gravity, typically \( 9.8 \; \text{m/s}^2 \).
Kinetic friction acts to oppose the motion of the block, taking some of the energy out of the system as the block slides. As seen in our calculation, this friction causes a loss of some of the block's energy, illustrated by the negative sign in the work done by friction.
Work-Energy Principle
The Work-Energy Principle is a staple concept in physics. It states that the work done on an object is equal to the change in its kinetic energy. When dealing with scenarios involving springs and friction, this principle is used in conjunction with the Conservation of Energy principle.

In our problem, we consider various types of energy:- **Initial Elastic Potential Energy**: The energy stored initially in the spring.- **Work Done by Friction**: The energy loss due to friction as the block moves.- **Final Elastic Potential Energy**: The energy remaining in the spring at a lesser compression.- **Final Kinetic Energy**: The energy indicative of the block鈥檚 speed as it moves.The main equation reads:\[(PE_{initial} + W_{friction}) = (KE_{final} + PE_{final})\]By rearranging and calculating each part, we find the speed of the block after moving a particular distance. This exercise effectively shows how energy is transferred and transformed within a physical system.

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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 -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 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.

An ideal spring of negligible mass is 12.00 \(\mathrm{cm}\) long when nothing is attached to it. When you hang a 3.15 -kg weight from it, you measure its length to be 13.40 \(\mathrm{cm}\) . If you wanted to store 10.0 \(\mathrm{J}\) of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.

A hydroelectric dam holds back a lake of surface area \(3.0 \times 10^{6} \mathrm{m}^{2}\) that has vertical sides below the water level. The water level in the lake is 150 \(\mathrm{m}\) above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is converted to electrical energy with 90\(\%\) efficiency. (a) If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the water in the lake? The density of water is 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) . (b) What volume of water must pass through the dam to produce 1000 kilowatt-hours of electrical energy? What distance does the level of water in the lake fall when this much water passes through the dam?

A 60.0 -kg skier sturts from rest at the top of a ski slope 65.0 \(\mathrm{m}\) high. (a) If frictional forces do \(-10.5 \mathrm{kJ}\) of work on her as she descends, how fast is she going at the bottom of the slope? (b) Now moving horizontally, the skier crosses a patch of soft snow, where \(\mu_{\mathrm{k}}=0.20 .\) If the patch is 82.0 \(\mathrm{m}\) wide and the average force of air resistance on the skier is 160 \(\mathrm{N}\) , how fast is she going after crossing the patch? (c) The skier hits a snowdrift and penetrates 2.5 \(\mathrm{m}\) into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?

CP A 120 -kg mail bag hangs by a vertical rope 3.5 \(\mathrm{m}\) long. A postal worker then displaces the bag to a position 2.0 \(\mathrm{m}\) sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?
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