/*! 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 21 A block of mass \(1 \mathrm{~kg}... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 21

A block of mass \(1 \mathrm{~kg}\) kept on a rough horizontal surface \((u=0.4)\) is attached to a light spring (force constant \(=200 \mathrm{~N} / \mathrm{m}\) ) whose other end is attached to a vertical wall. The block is pushed to compress the spring by a distance \(d\) and released. Find the value(s) of ' \(d\) ' for which (spring + block) system loses its entire mechanical energy in form of heat. (1) \(4 \mathrm{~cm}\) (2) \(6 \mathrm{~cm}\) (3) \(8 \mathrm{~cm}\) (4) \(10 \mathrm{~cm}\)

Short Answer

Expert verified
The system loses its entire mechanical energy as heat for \(d = 4 \, \text{cm}\). (Option 1)

Step by step solution

01

Identify the Concept

To find the distance \(d\) for which the entire mechanical energy is converted to heat, we need to equate the work done against friction to the initial potential energy stored in the spring. This condition signifies the total mechanical energy loss as heat.
02

Calculate Spring Potential Energy

The potential energy stored in the spring when compressed by distance \(d\) is given by: \[ U = \frac{1}{2} k d^2 \] where \(k = 200 \, \text{N/m}\) is the spring constant.
03

Calculate Work Done Against Friction

The work done against friction is given by: \[ W = f_k \cdot d \] where \( f_k = \mu \cdot m \cdot g \) is the frictional force. Given \( \mu = 0.4, \; m = 1 \, \text{kg}, \; g = 9.8 \, \text{m/s}^2 \), we find \( f_k = 0.4 \cdot 1 \cdot 9.8 = 3.92 \; \text{N} \). Thus, \( W = 3.92d \).
04

Equate Energy Loss to Work Done

Set the spring potential energy equal to the work done by friction: \[ \frac{1}{2} \times 200 \times d^2 = 3.92 \times d \] Simplify and solve for \(d\): \[ 100d^2 = 3.92d \] \[ 100d = 3.92 \] \[ d = \frac{3.92}{100} \] \[ d = 0.0392 \, \text{m} = 3.92 \, \text{cm} \]
05

Match Result with Given Choices

Now, compare the calculated \(d = 3.92 \, \text{cm}\) with the given options. The closest value to the choices is \(4 \, \text{cm}\). This is the choice for which the system loses entire mechanical energy as heat.

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Key Concepts

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

Spring Potential Energy
The spring potential energy is the stored energy in a compressed or stretched spring, resulting from its deformation. It's essentially the energy associated with the position of the spring. When a force compresses or extends the spring, it accumulates potential energy to be released later. This energy is expressed mathematically as: \[ U = \frac{1}{2} k d^2 \] where:
  • U is the potential energy,
  • k represents the spring constant in N/m,
  • d is the distance compressed or stretched, measured in meters.
Understanding that the spring constant, k, indicates the stiffness of the spring, and how it affects the potential energy, is crucial. A stiffer spring with a higher spring constant will store more energy for the same deformation, compared to a less stiff spring. This formula underpins the conversion and transfer of energy within the spring-block system we are analyzing.
Work Done Against Friction
To understand the subsequent loss of energy, we must consider the work done against friction. As the block moves over the rough surface, it encounters frictional force, which resists motion. This frictional force (fk) is calculated as: \[ fk = \mu \cdot m \cdot g \] where:
  • \mu is the coefficient of friction,
  • m is the mass of the block, and
  • g is the acceleration due to gravity.
In our problem: \[ fk = 0.4 \cdot 1 \cdot 9.8 = 3.92 \, \text{N} \] The work done to overcome this friction as the block moves the distance d is given by: \[ W = fk \cdot d \] This work done is essentially the energy lost due to friction, converted entirely into heat. As the block moves, it dissipates energy, meaning that the energy originally stored in the spring is gradually converted into thermal energy.
Energy Conversion in Springs
The interaction between the spring and the block is a classic example of energy conversion. Initially, when the spring is compressed, energy is stored in it as spring potential energy. When released, this energy is converted into kinetic energy, initiating the movement of the block.
As the block slides, it encounters friction, which transforms kinetic energy into thermal energy鈥攅nergy lost in the system, often considered unwanted in mechanical contexts. The role of energy conversion in systems like these is to illustrate how energy changes form but is conserved overall.
In our system where energy is entirely lost to friction, calculating the point where spring potential energy equals the work done against friction helps determine when all mechanical energy is lost as heat. The law of conservation of energy holds, affirming the total energy initially stored in the spring equals the energy expended against friction. This conversion illustrates the transient journey of energy through various states during mechanical processes.

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Most popular questions from this chapter

A particle is taken from point \(A\) to point \(B\) under the influence of a force field. Now it is taken back from \(B\) to \(A\) and it is observed that the work done in taking the particle from \(A\) to \(B\) is not equal to the work done in taking it from \(B\) to \(A\). If \(W_{n c}\) and \(W_{c}\) are the work done by non- conservative and conservative forces present in the system, respectively \(\Delta U\) is the change in potential energy and \(\Delta k\) is the chan in kinetic energy, then (1) \(W_{n c}-\Delta U=\Delta k\) (2) \(W_{e}=-\Delta U\) (3) \(W_{n c}^{n}+W_{c}=\Delta k\) (4) \(W_{n c}-\Delta U=-\Delta k\)

A \(1.5-\mathrm{kg}\) block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of \(x\)-axis is applied to the block. The force is given by \(\vec{F}=\left(4-x^{2}\right) \vec{i} \mathrm{~N}\), where \(x\) is in meter and the initial position of the block is \(x=0\). The maximum kinetic energy of the block between \(x\) and \(x=2.0 \mathrm{~m}\) is (1) \(2.33 \mathrm{~J}\) (2) \(8.67 \mathrm{~J}\) (3) \(5.33 \mathrm{~J}\) (4) \(6.67 \mathrm{~J}\)

Let \(r\) be the distance of a particle from a fixed point which it is attracted by an inverse square law force given \(F=k / r^{2}(k=\) constant \() .\) Let \(m\) be the mass of the partic and \(L\) be its angular momentum with respect to the fixe point. Which of the following formulae is correct about th total energy of the system? (1) \(\frac{1}{2} m\left(\frac{d r}{d t}\right)^{2}-\frac{k}{r}+\frac{L}{2 m r^{2}}=\) Constant (2) \(\frac{1}{2} m\left(\frac{d r}{d t}\right)^{2}-\frac{k}{r}=\) Constant (3) \(\frac{1}{2} m\left(\frac{d r}{d t}\right)^{2}+\frac{k}{r}+\frac{L^{2}}{2 m r^{2}}=\) Constant (4) None

A block of mass \(2 \mathrm{~kg}\) is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force \(F=40\) \(\mathrm{N}\). The kinetic energy of the particle increase \(40 \mathrm{~J}\) in a given interval of time. Then: \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (1) tension in the string is \(40 \mathrm{~N}\) (2) displacement of the block in the given interval of time is \(2 \mathrm{~m}\) (3) work done by gravity is \(-20 \mathrm{~J}\) (4) work done by tension is \(80 \mathrm{~J}\)

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