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Chapter 7: Problem 51

(I) \(( a )\) If the kinetic energy of a particle is tripled, by what factor has its speed increased? (b) If the speed of a particle is halved, by what factor does its kinetic energy change?

Short Answer

Expert verified
(a) The speed increases by a factor of \( \sqrt{3} \). (b) The kinetic energy changes by a factor of \( \frac{1}{4} \).

Step by step solution

01

Understanding the Kinetic Energy Formula

Kinetic energy is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the speed of the particle.
02

Analyzing Part (a) - Tripling Kinetic Energy

If the kinetic energy is tripled, we replace \( KE \) with \( 3KE \). So, we have \( 3KE = \frac{1}{2}m(v')^2 \), where \( v' \) is the new speed.
03

Relating New KE to Original KE (a)

Since \( KE = \frac{1}{2}mv^2 \) and \( 3KE = \frac{1}{2}m(v')^2 \), set them equal to find \( v' \):\[ 3 \left( \frac{1}{2}mv^2 \right) = \frac{1}{2}m(v')^2 \] Simplifying, \( 3v^2 = (v')^2 \).
04

Solving for New Speed (a)

Take the square root of both sides: \( v' = \sqrt{3}v \). Thus, the speed increases by a factor of \( \sqrt{3} \).
05

Analyzing Part (b) - Halving Speed

If the speed is halved, we let \( v' = \frac{v}{2} \). Substituting into the KE formula: \( KE' = \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{2}m\frac{v^2}{4} = \frac{1}{4}KE \).
06

Finding Factor of Change in Kinetic Energy (b)

Since the new kinetic energy \( KE' = \frac{1}{4}KE \), the kinetic energy changes by a factor of \( \frac{1}{4} \).

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Key Concepts

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

Understanding the Kinetic Energy Formula
Kinetic energy is the energy a body possesses due to its motion. The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \), where \( m \) represents mass, and \( v \) indicates speed. This formula reveals that kinetic energy depends on both mass and speed. Particularly, it shows that kinetic energy is directly proportional to the square of the speed. This means when speed changes, kinetic energy changes even more dramatically. For example, if the speed doubles, the kinetic energy increases by a factor of four. Understanding this formula is crucial for solving problems related to motion and energy changes.
Speed and Energy Relationship
The relationship between speed and kinetic energy is profound. Since kinetic energy is proportional to the square of speed, even small changes in speed lead to more significant changes in kinetic energy. Let's explore this with some examples:
  • If you increase the speed, the kinetic energy goes up much faster than the speed itself.
  • Tripling the speed, for instance, will result in a ninefold increase in kinetic energy because \((3v)^2 = 9v^2\).
  • On the other hand, if you reduce the speed, the kinetic energy decreases rapidly.
  • Halving the speed will cut the kinetic energy down to a quarter, since \(\left( \frac{v}{2} \right)^2 = \frac{v^2}{4}\).
Understanding this relationship helps to predict how varying speeds impact the energy of a moving object.
Tripling and Halving Concepts
Let's delve into what happens to kinetic energy when speed is tripled or halved. These operations help illustrate the formula's implications.
Consider tripling the kinetic energy of an object. To find the new speed, we set up the equation \( 3KE = \frac{1}{2}m(v')^2 \) and compare it to the original \( KE = \frac{1}{2}mv^2 \). Solving \( 3v^2 = (v')^2 \), we discover \( v' = \sqrt{3}v \). In simple terms, to achieve three times the kinetic energy, the speed must increase by a factor of \( \sqrt{3} \).
Now, consider what happens when the speed itself is halved. The speed change reflects as \( v' = \frac{v}{2} \), making the new kinetic energy \( KE' = \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{4}KE \). Thus, halving the speed results in kinetic energy reducing to a quarter of its original value.
These concepts are essential in understanding the exponential nature of changes in kinetic energy relative to changes in speed.

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Most popular questions from this chapter

(III) We usually neglect the mass of a spring if it is small compared to the mass attached to it. But in some applications, the mass of the spring must be taken into account. Consider a spring of unstretched length \(\ell\) and mass \(M _ { \text { s uniformly } }\) distributed along the length of the spring. A mass \(m\) is attached to the end of the spring. One end of the spring is fixed and the mass \(m\) is allowed to vibrate horizontally without friction (Fig. \(30 ) .\) Each point on the spring moves with a velocity proportional to the distance from that point to the fixed end. For example, if the mass on the end moves with speed \(v _ { 0 } ,\) the midpoint of the spring moves with speed \(v _ { 0 } / 2 .\) Show that the kinetic energy of the mass plus spring when the mass is moving with velocity \(v\) is \(K = \frac { 1 } { 2 } M v ^ { 2 }\) where \(M = m + \frac { 1 } { 3 } M _ { \mathrm { S } }\) is the "effective mass" of the system. [Hint: Let \(D\) be the total length of the stretched spring. Then the velocity of a mass \(d m\) of a spring of length \(d x\) located at \(x\) is \(v ( x ) = v _ { 0 } ( x / D ) .\) Note also that \(d m = d x \left( M _ { \mathrm { S } } / D \right) . ]\)

\((a)\) If the kinetic energy of a particle is tripled, by what factor has its speed increased? (b) If the speed of a particle is halved, by what factor does its kinetic energy change?

A force \(\mathbf { F } = ( 10.0 \mathbf { i } + 9.0 \mathbf { j } + 12.0 \mathbf { k } ) \mathrm { kN }\) acts on a small object of mass 95\(\mathrm { g } .\) If the displacement of the object is \(\mathbf { d } = ( 5.0 \mathbf { i } + 4.0 \mathbf { j } ) \mathrm { m } ,\) find the work done by the force. What is the angle between \(\vec { \mathbf { F } }\) and \(\mathbf { d } ?\)

A 3.0 -m-long steel chain is stretched out along the top level of a horizontal scaffold at a construction site, in such a way that \(2.0 \mathrm{~m}\) of the chain remains on the top level and \(1.0 \mathrm{~m}\) hangs vertically, Fig. \(7-26 .\) At this point, the force on the hanging segment is sufficient to pull the entire chain over the edge. Once the chain is moving, the kinetic friction is so small that it can be neglected. How much work is performed on the chain by the force of gravity as the chain falls from the point where \(2.0 \mathrm{~m}\) remains on the scaffold to the point where the entire chain has left the scaffold? (Assume that the chain has a linear weight density of \(18 \mathrm{~N} / \mathrm{m}\).)

What is the minimum work needed to push a \(950-\mathrm{kg}\) car \(310 \mathrm{~m}\) up along a \(9.0^{\circ}\) incline? Ignore friction.
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