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Chapter 7: Problem 51

\((a)\) 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) Speed increases by \( \sqrt{3} \); (b) Kinetic energy decreases by 4.

Step by step solution

01

Understanding Kinetic Energy Equation

The kinetic energy (KE) of a particle is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the speed of the particle. We need to determine how a change in KE affects speed and vice versa.
02

Analyzing Part (a)

For part (a), if the kinetic energy is tripled, let the initial kinetic energy be \( KE_1 = \frac{1}{2}mv_1^2 \). The new kinetic energy is \( KE_2 = 3KE_1 = \frac{1}{2}mv_2^2 \). Simplifying gives \( 3 \times \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 \).
03

Solving for Speed in Part (a)

Cancelling out the \( \frac{1}{2}m \) from both sides in \( 3 \times \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 \) yields \( 3v_1^2 = v_2^2 \). Taking the square root gives \( v_2 = \sqrt{3}v_1 \). Thus, the speed increases by a factor of \( \sqrt{3} \).
04

Analyzing Part (b)

For part (b), if the speed is halved, let the initial speed be \( v_1 \). The new speed is \( v_2 = \frac{1}{2}v_1 \). The initial kinetic energy is \( KE_1 = \frac{1}{2}mv_1^2 \) and the new kinetic energy is \( KE_2 = \frac{1}{2}m\left(\frac{1}{2}v_1\right)^2 \).
05

Solving for Kinetic Energy in Part (b)

Substitute \( v_2 \) into the kinetic energy equation: \( KE_2 = \frac{1}{2}m\left(\frac{1}{2}v_1\right)^2 = \frac{1}{2}m\frac{1}{4}v_1^2 = \frac{1}{4} \times \frac{1}{2}mv_1^2 \). Thus, \( KE_2 = \frac{1}{4}KE_1 \), showing that the kinetic energy is reduced by a factor of 4.

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Key Concepts

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

Kinetic Energy Equation
Understanding the kinetic energy equation is crucial for solving any problem related to kinetic energy. Kinetic energy (\( KE \)) of a particle is calculated as \( KE = \frac{1}{2}mv^2 \), where \( m \) represents the mass and \( v \) represents speed. This formula shows that kinetic energy depends on two main factors:
  • Mass (\( m \))
  • Speed (\( v \))
Since speed is squared, it has a more significant impact on the kinetic energy than the mass does. For a particle with doubled speed, its kinetic energy increases by a factor of four, given mass remains constant.
This equation will help us understand how changes in speed affect kinetic energy and vice versa.
Speed and Kinetic Energy Relationship
The relationship between speed and kinetic energy is quite fascinating. When addressing how changes in kinetic energy affect speed, it helps to know the direct relationship:
  • When kinetic energy increases, the speed of a particle also increases.
  • The relationship is quadratic, meaning a slight increase in speed results in a significant change in kinetic energy.
In the previously solved exercise, if the kinetic energy triples, we determined that the speed increases by the square root of three. Using the kinetic energy formula \( KE_2 = 3 \times \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 \), we canceled terms and solved \( v_2 = \sqrt{3}v_1 \).
Conversely, if speed is halved, the relationship shows the kinetic energy is reduced by a factor of four due to the squared speed term.
Factors of Speed and Energy Change
Understanding how different factors affect speed and energy change allows for deeper insight into kinetic phenomena. Key points to consider:
  • Mass: Changes in mass do not affect this particular relationship since mass is constant in our exercise, focusing primarily on speed changes.
  • Speed: Plays a crucial role in determining kinetic energy. Small changes in speed result in exponential changes in energy.
From the exercise, when speed is halved, the energy decreased fourfold, demonstrating how significant speed changes directly affect kinetic energy. Examining various scenarios helps in understanding how each element impacts both speed and kinetic energy. Recognizing these changes assists students in a broader range of applications across physics problems.

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Most popular questions from this chapter

A \(75-\mathrm{kg}\) meteorite buries itself \(5.0 \mathrm{~m}\) into soft mud. The force between the meteorite and the mud is given by \(F(x)=\left(640 \mathrm{~N} / \mathrm{m}^{3}\right) x^{3},\) where \(x\) is the depth in the mud. What was the speed of the meteorite when it initially impacted the mud?

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 \(M_{\mathrm{S}}\) 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. \(7-30\) ). 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 $$ d m=d x\left(M_{\mathrm{S}} / D\right) $$.

If \(\overrightarrow{\mathbf{A}}=9.0 \hat{\mathbf{i}}-8.5 \hat{\mathbf{j}}, \quad \overrightarrow{\mathbf{B}}=-8.0 \hat{\mathbf{i}}+7.1 \hat{\mathbf{j}}+4.2 \hat{\mathbf{k}},\) and \(\overrightarrow{\mathbf{C}}=6.8 \hat{\mathbf{i}}-9.2 \hat{\mathbf{j}},\) determine \((a) \overrightarrow{\mathbf{A}} \cdot(\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}) ;(b)(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{C}}) \cdot \overrightarrow{\mathbf{B}} ;\) \((c)(\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{A}}) \cdot \mathbf{C}\)

A varying force is given by \(F = A e ^ { - k x } ,\) where \(x\) is the position; \(A\) and \(k\) are constants that have units of \(\mathrm { N }\) and \(\mathrm { m } ^ { - 1 }\) , respectively. What is the work done when \(x\) goes from 0.10\(\mathrm { m }\) to infinity?

An airplane pilot fell 370\(\mathrm { m }\) after jumping from an aircraft without his parachute opening. He landed in a snowbank, creating a crater 1.1\(\mathrm { m }\) deep, but survived with only minor injuries. Assuming the pilot's mass was 88\(\mathrm { kg }\) and his terminal velocity was 45\(\mathrm { m } / \mathrm { s }\) , estimate: \(( a )\) the work done by the snow in bringing him to rest; (b) the average force exerted on him by the snow to stop him; and (c) the work done on him by air resistance as he fell. Model him as a particle.
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