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A particle of mass \(2 \mathrm{~kg}\) starts moving in a straight line with an initial velocity of \(2 \mathrm{~m} / \mathrm{s}\) at a constant acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\). The rate of change of kinetic energy is (A) four times the velocity at any moment. (B) two times the displacement at any moment. (C) four times the rate of change of velocity at any moment. (D) constant throughout.

Short Answer

Expert verified
The correct answer is (C): the rate of change of kinetic energy is four times the rate of change of velocity at any moment. The rate of change of kinetic energy as a function of time is \( \frac{dK}{dt} = 16 + 16t \).

Step by step solution

01

Define given values and equations

We are given: Mass (m) = 2 kg Initial velocity (u) = 2 m/s Acceleration (a) = 2 m/s² The equations of motion are: 1) v = u + at 2) s = ut + (1/2)at² 3) v² = u² + 2as Additionally, the kinetic energy (K) is given by the formula: \( K = \frac{1}{2}mv^2 \)
02

Find the velocity as a function of time

To find the velocity as a function of time, we use equation (1): \( v(t) = u + at \) Substitute the given values: \( v(t) = 2 + 2t \)
03

Substitute the velocity function into the kinetic energy formula

Now, let's substitute the velocity function into the kinetic energy formula: \( K(t) = \frac{1}{2}m(v(t))^2 \) \( K(t) = \frac{1}{2}(2)(2 + 2t)^2 \)
04

Calculate the rate of change of kinetic energy

To find the rate of change of kinetic energy, we need to differentiate the kinetic energy equation with respect to time: \( \frac{dK}{dt} = \frac{d}{dt} \big[ (2)((2 + 2t)^2) \big] \) We can simplify the expression inside the square bracket: \( \frac{dK}{dt} = \frac{d}{dt} \big[ (2)(4 + 8t + 4t^2) \big] \) Now, differentiate with respect to time: \( \frac{dK}{dt} = 2(8 + 8t) \)
05

Determine the correct option

We have found the rate of change of kinetic energy as a function of time: \( \frac{dK}{dt} = 16 + 16t \) Now, let's examine the options: (A) four times the velocity at any moment. (B) two times the displacement at any moment. (C) four times the rate of change of velocity at any moment. (D) constant throughout. (A) is not correct, as the rate of change of kinetic energy is not 4 times the velocity function (which is 2+2t). (B) is not correct, as it does not correlate with displacement. (C) Since the rate of change of velocity is equal to the acceleration (which is 2 m/s²), we can rewrite the rate of change of kinetic energy as: \( \frac{dK}{dt} = 8(2) \) So, the correct answer is (C): four times the rate of change of velocity at any moment. (D) is not correct, as we see that the rate of change of kinetic energy is not constant.

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Key Concepts

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

Kinetic Energy in Physics
When a student dives into the study of mechanics, one of the first and most important concepts they encounter is kinetic energy. This term represents the energy that an object possesses due to its motion. The kinetic energy equation,
\( K = \frac{1}{2}mv^2 \)
,where \( m \) is the mass and \( v \) is the velocity of the object, underpins much of classical mechanics, facilitating the analysis of systems ranging from simple pendulums to complex vehicles. Understanding that the kinetic energy is directly proportional to the square of the velocity reveals why high-speed impacts are dramatically more energetic than low-speed contacts.
For the improving student, a crucial application of kinetic energy is its role in work-energy principles, where we often calculate the work done on a body by realizing it's actually a change in its kinetic energy.
Equations of Motion
The equations of motion are foundational tools for any physicist. They describe how an object's velocity and displacement evolve over time under the influence of constant acceleration. The three key equations are:

First Equation of Motion:

\( v = u + at \)
, where \( v \) is the final velocity, \( u \) is the initial velocity, and \( a \) is the acceleration.

Second Equation of Motion:

\( s = ut + \frac{1}{2}at^2 \)
, \( s \) being the displacement after time \( t \).

Third Equation of Motion:

\( v^2 = u^2 + 2as \)
.
These equations perfectly illustrate the link between an object's kinematics and the forces applied to it. They are crucial for problems dealing with constant acceleration, such as vehicles accelerating down a road or objects falling under the force of gravity. For students, mastering these equations opens the door to solving many real-world kinematics problems.
Differentiation in Kinematics
In physics, and particularly in kinematics, differentiation is a mathematical operation that provides the rate of change of a function. When velocity is the function of time, its differentiation gives acceleration. Conversely, when we have a function for position, differentiating it with respect to time gives velocity.
Taking it further, the differentiation of velocity with respect to time expresses acceleration, while differentiating the kinetic energy function reveals the rate of power or the rate at which work is being done, which is deeply connected to the concept of force. This relationship can be seen in the exercise where the differentiation of kinetic energy with respect to time gives the rate of change of kinetic energy:
\( \frac{dK}{dt} \)
.For learners, the realization that these concepts are not just theoretical but are constantly at play in everyday phenomena enhances their relevance. It illustrates how mathematical concepts like differentiation directly apply to the physical world and empowers students with tools to analyse and predict the behaviour of dynamic systems.

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Most popular questions from this chapter

Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. Then, if the spheres roll without slipping. (A) the heavier sphere reaches the bottom first. (B) the bigger sphere reaches the bottom first. (C) the two spheres reach the bottom together. (D) velocity of the both sphere will be same at the bottom.

A body of mass \(m_{1}\) makes a head on collision perfectly elastic with a body of mass \(m_{2}\) initially at rest. (i) What fraction of initial energy of mass \(m_{1}\) is lost in collision? (ii) For what ratio of \(\frac{m_{2}}{m_{1}}=\eta\), the fraction of energy loss is maximum?

An isolated particle of mass \(\mathrm{m}\) is moving in horizontal plane \((x-y)\), along the \(x\)-axis, at a certain height above ground. It suddenly explodes into two fragments of masses \(\mathrm{m} / 4\) and \(3 \mathrm{~m} / 4\). An instant later, the smaller fragment is at \(y=+15 \mathrm{~cm}\). The larger fragment at this instant is at (A) \(y=-5 \mathrm{~cm}\) (B) \(y=+5 \mathrm{~cm}\) (C) \(y=+5 \mathrm{~cm}\) (D) \(y=-20 \mathrm{~cm}\)

Distance of the centre of mass of a solid uniform cone from its vertex is \(z_{0}\). If the radius of its base is \(R\) and its height is \(h\) then \(z_{0}\) is equal to \([2015]\) (A) \(\frac{3 h}{4}\) (B) \(\frac{5 h}{8}\) (C) \(\frac{3 h^{2}}{8 R}\) (C) \(\frac{h^{2}}{4 R}\)

A particle of mass \(m\) describes a circle of radius \(r\). The centripetal acceleration of the particle is \(4 / r^{2}\). The momentum of the particle is (A) \(\frac{4 m}{r}\) (B) \(\frac{2 m}{r}\) (C) \(\frac{4 m}{\sqrt{r}}\) (D) \(v\)

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