
Derive the expression for K.E. of mass $m$ moving with velocity $\left( v \right)$.
Answer
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Hint: To solve this question, we need to consider the given mass to be at rest on which a given force acts due to which the mass attains the given velocity. Then we have to find out the work done on the particle due to the force. Finally, using the second law of motion, and the third kinematic equation of motion, we can get the work done in terms of the given parameters, which will be the required expression for the kinetic energy.
Formula used:
The formula used to solve this question is given by
$W = F \cdot s$, $W$ is the work done by a force $F$ in displacing a particle through a displacement of $s$.
${v^2} - {u^2} = 2as$, here $u$ is the initial velocity, $v$ is the final velocity, $a$ is the acceleration, and $s$ is the displacement.
Complete step-by-step answer:
Consider a particle of mass $m$ which is at rest.
Let a force of $F$ acts on the particle for some time due to which it attains a velocity of $v$. In this time, let the particle be displaced by a displacement $s$.
We know that the work done is given by
$W = F \cdot s$
As the force and the displacement of the particle are in the same direction, so we have
$W = Fs$ …………………...(1)
Now, from the thirds kinematic equation, we have
${v^2} - {u^2} = 2as$
As the particle is initially at rest, so we have $u = 0$. Substituting it above, we get
${v^2} = 2as$
$ \Rightarrow s = \dfrac{{{v^2}}}{{2a}}$ ………………….(2)
Substituting (2) in (1) we get
$W = F\left( {\dfrac{{{v^2}}}{{2a}}} \right)$
$ \Rightarrow W = \dfrac{F}{a}\left( {\dfrac{{{v^2}}}{2}} \right)$ ……………….(3)
From the Newton’s second law of motion, we know that
$F = ma$ …………………………..(4)
Substituting (4) in (3) we get
$W = \dfrac{{ma}}{a}\left( {\dfrac{{{v^2}}}{2}} \right)$
$ \Rightarrow W = \dfrac{{m{v^2}}}{2}$
This work done by the force is stored as the kinetic energy of the particle. So the kinetic energy of the particle of mass $m$ which is moving with a velocity of $v$ is given by
$K = \dfrac{{m{v^2}}}{2}$
Hence, this is the required expression for the kinetic energy of the given mass.
Note: In this derivation, we have assumed the force to be constant. This assumption was necessary so that the acceleration of the particle became uniform and the third kinematic equation could be applied.
Formula used:
The formula used to solve this question is given by
$W = F \cdot s$, $W$ is the work done by a force $F$ in displacing a particle through a displacement of $s$.
${v^2} - {u^2} = 2as$, here $u$ is the initial velocity, $v$ is the final velocity, $a$ is the acceleration, and $s$ is the displacement.
Complete step-by-step answer:
Consider a particle of mass $m$ which is at rest.
Let a force of $F$ acts on the particle for some time due to which it attains a velocity of $v$. In this time, let the particle be displaced by a displacement $s$.
We know that the work done is given by
$W = F \cdot s$
As the force and the displacement of the particle are in the same direction, so we have
$W = Fs$ …………………...(1)
Now, from the thirds kinematic equation, we have
${v^2} - {u^2} = 2as$
As the particle is initially at rest, so we have $u = 0$. Substituting it above, we get
${v^2} = 2as$
$ \Rightarrow s = \dfrac{{{v^2}}}{{2a}}$ ………………….(2)
Substituting (2) in (1) we get
$W = F\left( {\dfrac{{{v^2}}}{{2a}}} \right)$
$ \Rightarrow W = \dfrac{F}{a}\left( {\dfrac{{{v^2}}}{2}} \right)$ ……………….(3)
From the Newton’s second law of motion, we know that
$F = ma$ …………………………..(4)
Substituting (4) in (3) we get
$W = \dfrac{{ma}}{a}\left( {\dfrac{{{v^2}}}{2}} \right)$
$ \Rightarrow W = \dfrac{{m{v^2}}}{2}$
This work done by the force is stored as the kinetic energy of the particle. So the kinetic energy of the particle of mass $m$ which is moving with a velocity of $v$ is given by
$K = \dfrac{{m{v^2}}}{2}$
Hence, this is the required expression for the kinetic energy of the given mass.
Note: In this derivation, we have assumed the force to be constant. This assumption was necessary so that the acceleration of the particle became uniform and the third kinematic equation could be applied.
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