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Chapter 3: Problem 52

The rate of change of kinetic energy is "n" times ( \(n\) is constant with appropriate dimension) the velocity at any moment of a particle of mass \(\mathrm{m}\), which starts moving in a straight line. The constant acceleration with which particle is moving is : (1) \(\mathrm{mn}\) (2) \(\frac{\mathrm{m}}{\mathrm{n}}\) (3) \(\frac{\mathrm{n}}{\mathrm{m}}\) (4) \(\frac{1}{\mathrm{mn}}\)

Short Answer

Expert verified
The constant acceleration is \( \frac{n}{m} \), which corresponds to option (3).

Step by step solution

01

Understand Problem Context

Identify that the problem involves kinetic energy (KE) and its rate of change with respect to velocity for a particle of mass \(m\). Kinetic Energy (KE) is given by \( KE = \frac{1}{2} mv^2 \), where \(v\) is the velocity of the particle.
02

Define Rate of Change

The problem states that the rate of change of kinetic energy is given by the expression \( n \times v \). Mathematically, this can be represented as: \[ \frac{d(KE)}{dt} = n \times v \]
03

Differentiate Kinetic Energy

Differentiate \( KE = \frac{1}{2} mv^2 \) with respect to time \(t\). This gives: \[ \frac{d(KE)}{dt} = \frac{1}{2} m \frac{d(v^2)}{dt} \] Using the chain rule: \[ \frac{d(KE)}{dt} = \frac{1}{2} m \cdot 2v \cdot \frac{dv}{dt} = mv \frac{dv}{dt} \]
04

Compare Expressions

Compare the differentiated expression \( mv \frac{dv}{dt} \) with the given rate of change expression \( n \times v \). Setting them equal, we have: \[ mv \frac{dv}{dt} = nv \]
05

Solve for Acceleration

Solve for \( \frac{dv}{dt} \), which is the acceleration \(a\): \[ mv \frac{dv}{dt} = nv \] Dividing both sides by \(v\): \[ m \frac{dv}{dt} = n \] Therefore, the acceleration \(a = \frac{dv}{dt} = \frac{n}{m} \)
06

Identify Correct Answer

From the solved acceleration \(a = \frac{n}{m} \), the correct option is (3).

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Key Concepts

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

kinetic energy
Kinetic energy (KE) is the energy that a particle has due to its motion. For a particle of mass \(m\) moving with a velocity \(v\), the kinetic energy is given by the formula: \ \ \( KE = \frac{1}{2} mv^2 \). This equation shows that the kinetic energy is directly proportional to the mass of the particle and the square of its velocity. As the velocity increases, the kinetic energy increases quadratically. \ \ For example, if a particle's velocity doubles, its kinetic energy increases by a factor of four. This relationship highlights the significant impact velocity has on the kinetic energy of a particle.
differentiation
Differentiation is a crucial concept in calculus that involves finding the rate at which a quantity changes. In this exercise, we need to differentiate the kinetic energy with respect to time. \ \ The differentiation of kinetic energy with time helps us understand how quickly the energy changes as the particle's velocity changes. Given \( KE = \frac{1}{2} mv^2 \), we apply differentiation to the equation to get: \ \ \( \frac{d(KE)}{dt} = \frac{1}{2} m \frac{d(v^2)}{dt} \). \ \ This step is necessary to connect how kinetic energy changes with time to the particle's velocity and acceleration.
acceleration calculation
Acceleration is the rate of change of velocity with respect to time. In this problem, after differentiating kinetic energy, we aim to find the particle's constant acceleration. \ \ First, we observe that the rate of change of kinetic energy is given by \( n \times v \): \ \ \( \frac{d(KE)}{dt} = n \times v \). \ \ Using our differentiated kinetic energy expression, we set them equal: \ \ \( mv \frac{dv}{dt} = nv \). \ \ Simplifying this by dividing both sides by \(v\), we get: \ \ \( m \frac{dv}{dt} = n \). \ \ Therefore, the acceleration \(a\) is \( \frac{dv}{dt} = \frac{n}{m} \).
chain rule
The chain rule is a fundamental tool in differentiation used to find the derivative of a composite function. In this exercise, we apply the chain rule to differentiate \(v^2\) with respect to time \(t\). \ \ The chain rule states that \( \frac{d}{dt} [f(g(t))] = f'(g(t)) \cdot g'(t) \). For our problem, let \( f(u) = \frac{1}{2} mu \) and \( u = v^2 \). Therefore, \ \ \( \frac{d(KE)}{dt} = \frac{1}{2} m \cdot 2v \cdot \frac{dv}{dt} = mv \frac{dv}{dt} \). \ \ This application of the chain rule simplifies the differentiation process and connects changes in velocity to changes in kinetic energy, making it a powerful technique in calculus.

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