/*! 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 114 A body of mass \(m\) travels in ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 114

A body of mass \(m\) travels in a straight line with a velocity \(v=k x^{3 / 2}\) where \(k\) is a constant. The work done in displacing the body from \(x=0\) to \(x\) is proportional to (a) \(x^{1 / 2}\) (b) \(x^{2}\) (c) \(x^{3}\) (d) \(x^{5 / 2}\)

Short Answer

Expert verified
The work done is proportional to \(x^3\), so the correct option is (c).

Step by step solution

01

Express Kinetic Energy

The kinetic energy (KE) for a body with mass \(m\) and velocity \(v\) is given by the formula: \( KE = \frac{1}{2} m v^2 \). Since the velocity \(v\) is given as \(k x^{3/2}\), substitute this into the kinetic energy equation: \( KE = \frac{1}{2} m (k x^{3/2})^2 \). Simplify this to get \( KE = \frac{1}{2} m k^2 x^3 \).
02

Define Work Done

The work done \(W\) in moving the body from position \(x = 0\) to a position \(x = x\) is equal to the change in kinetic energy, since the work-energy principle states that the work done is equal to the change in kinetic energy of the body. Initially, at \(x = 0\), the kinetic energy is zero. Therefore, all work done is used to increase the kinetic energy to \(\frac{1}{2} m k^2 x^3\). Hence, \( W = \frac{1}{2} m k^2 x^3 - 0 \).
03

Simplify and Identify Proportionality

From Step 2, we see that \( W = \frac{1}{2} m k^2 x^3 \). Since \(m\), \(k\), and \(\frac{1}{2}\) are constants, \(W\) is directly proportional to \(x^3\). Therefore, the work done is proportional to \(x^3\). This means the correct option is (c).

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Key Concepts

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

Kinetic Energy
Kinetic energy is an essential concept in physics, describing the energy a body possesses due to its motion. When a body of mass \(m\) moves, its kinetic energy (KE) can be calculated using:
  • \( KE = \frac{1}{2} m v^2 \)
where \(v\) represents the velocity of the body. To gain deeper understanding, consider our given velocity, \(v = k x^{3/2}\), where \(k\) is a constant, and \(x\) signifies position. By substituting this velocity into the kinetic energy equation, we derive:
  • \( KE = \frac{1}{2} m (k x^{3/2})^2 \)
  • Simplifying further gives \( KE = \frac{1}{2} m k^2 x^3 \)
Hence, the kinetic energy is directly related to the position raised to the power of 3. This relationship later plays a crucial role in identifying how the work done is proportional to the displacement of the body.
Proportionality in Physics
In the realm of physics, understanding proportionality helps determine how one quantity changes in response to another. When we discuss the work done on a body, this often involves examining how it relates to changes in kinetic energy.From the problem, the velocity is given in terms of position, ultimately defining kinetic energy as \( KE = \frac{1}{2} m k^2 x^3 \). When we consider the work-energy principle:
  • Work done (W) equals the change in kinetic energy.
  • Initially, at \(x = 0\), the kinetic energy is zero.
  • The work done equals the kinetic energy at a position \(x\): \( W = KE \).
This simplifies to: \( W = \frac{1}{2} m k^2 x^3 \), indicating that the work done is directly proportional to \(x^3\). Thus, as the position increases, the work done follows a specific power dependency on \(x\), reflecting a clear proportionality.
Velocity-Position Relationship
The velocity-position relationship provides valuable insight into how the speed of an object varies with its location along a path. In the given problem, velocity \(v\) is expressed as a function of position \(x\): \(v = k x^{3/2}\).Understanding this function is key:
  • It implies that velocity isn鈥檛 constant but instead depends on how far the body has traveled (position).
  • As the object moves, \(x\) increases, resulting in changes to \(v\) in a non-linear manner, specifically \(x^{3/2}\).
  • Consequently, the kinetic energy transforms similarly since it involves \(v^2\), ultimately leading to \(x^3\) in its expression.
This non-linear relationship between velocity and position forms the foundation for understanding changes in kinetic energy and, subsequently, the proportional work done in moving the object from one position to another. Observing how velocity variably depends on position is crucial for grasping concepts within mechanics.

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Most popular questions from this chapter

If force is always perpendicular to motion: (a) kinctic energy remains constant (b) work done is zero (c) velocily is consuant (d) specd is constant

Choose the correct statement \((s)\) of the following: (a) Force acting on a particle for equal time intervals can produce the same change in momentum but different change in kinetic energy. (b) Force acting on a particle for equal displacements can produce the same change in kinetic energy but different change in momentum. (c) Force acting on a particle for equal time intervals can produce different change in momentum but the same change in kinetic energy. (d) Force acting on a particle for equal displacements can produce different change in kinetic energy but the same change in momentum.

Which of the following are correct? (a) A body moving with velocity \(v\) can be stopped over a distance \(s\). If the kinetic energy of the body is doubled the body can be stopped over distance \(2 s\), provided the retarding force remains unchanged. (b) The work done by a body is inversely proportional to time. (c) Work equal to \(16 \mathrm{~J}\) is done on a \(2 \mathrm{~kg}\) body to set it in motion. If whole of this work had been used up in increasing the kinetic energy, then the body would have acquired a velocity of \(4 \mathrm{~ms}^{-1}\). (d) If potential energy of a stretched spring is plotted against y-axis and (extension) \(^{2}\) against \(\mathrm{x}\) -axis, then the graph is a straight line.

The velocily of a particle moving along a line varics with distance as \(v=a \sqrt{x}\) where \(a\) is a constant. The work done by all forces when the particle moves from \(x=0\) to \(x=l\) is (mass of the particle is \(m\) ) (a) 0 (b) \(\mathrm{ma}^{2} l\) (c) \(\frac{1}{2} m a^{2} l\) (d) \(\frac{1}{3} m a^{2} l\)

The potential energy of a \(1 \mathrm{~kg}\) particle free to move along the \(x\) -axis is given by \(U(x)-\left(\frac{x^{4}}{4}-\frac{x^{2}}{2}\right) \mathrm{J} .\) The total mechanical energy of the particle is \(2 \mathrm{~J}\). What is the maximum specd of the particle?
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