Kinetic Energy

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Kinetic energy refers to the energy possessed by a body by virtue of its motion. Here are a few common examples of kinetic energy in everyday life:

The movement of air, which powers windmills.
The flow of water, which drives water mills.
The force of a hammer driving a nail into wood.
The ability of a bullet to penetrate a target because of its speed.

How Kinetic Energy is Measured

The kinetic energy of a body with mass $ M $ moving at velocity $ v $ can be determined in a couple of ways:

First method: By calculating the work done by the moving object against an external force until it comes to a stop.
Second method: By calculating the amount of work needed to accelerate the body from rest to a velocity $ v $.

Let’s determine the kinetic energy using the second method:

To calculate the kinetic energy of a body with mass $ M $ moving at velocity $ v $, imagine that the body starts from rest. A constant force $ \vec{F} $ is applied to the body, causing it to accelerate until it reaches a velocity $ v $ in the direction of the force. If $ \vec{S} $ represents the distance traveled during this time as shown in the figure, the work done by the force is:

$ \displaystyle W=\vec{F}\cdot \vec{S}=FS\ldots \text{(}\because \text{ }\theta ={{0}^{{}^\circ }})$

If $\vec{a}$ is the acceleration produced in the motion of the body, then according to Newton’s second law of motion:

$$ \vec{F} = M \vec{a} $$

As all the vectors $\vec{S}$, $\vec{v}$, and $\vec{a}$ are along the direction of $\vec{F}$, the arrowhead may be dropped and we can simply use their magnitudes. Therefore, equation (1.10) becomes:

$$ W=MaS \quad \ldots \text{(1)}$$$

Using the relation $v^2 – u^2 = 2 a S$, we have:

$$ v^2 – (0)^2 = 2 a S \quad \text{or} \quad a = \frac{v^2}{2S} $$

From equation (1.11), we have:

$$ W = M \frac{v^2}{2S} S = \frac{1}{2} M v^2 $$

As this much amount of work done on the body of mass $M$ will set the body of mass $M$ moving with velocity $v$ from its state of rest, it is equal to the kinetic energy of the body.

Therefore, the kinetic energy $= \dfrac{1}{2} M v^2 \quad \ldots \text{(2)}$

Alternative Method:

The expression for kinetic energy can also be derived using the calculus method as follows:

Suppose that a body is initially at rest and force $\vec{F}$ is applied on the body to displace it through a distance $d\vec{s}$ along its own direction. Then, small work done:

$$ dW = \vec{F} \cdot d\vec{s} = F ds \quad (\because \theta = 0^\circ) $$

According to Newton’s second law of motion:

$$ F = Ma $$

where $a$ is the acceleration produced (in the direction of force) by applying the force. Therefore,

$$ \begin{array}{l}dW=Mads\\=M\dfrac{{dv}}{{dt}}ds\ldots (\because a=\dfrac{{dv}}{{dt}})\end{array} $$

$$ dW = M \frac{ds}{dt} dv = M v dv \quad (\because \frac{ds}{dt} = v) $$

Therefore, the work done on the body in order to increase its velocity from zero to $v$ is given by:

$$ W = \int_0^v M v dv = M \left[ \frac{v^2}{2} \right]_0^v $$

$$ W = \frac{1}{2} M v^2 $$

This work done is equal to the kinetic energy of the body of mass $M$, when moving with velocity $v$.

Remember 1: In case the force applied increases the velocity from $u$ to $v$, then:

$$ W = \int_u^v M v dv = M \left[ \frac{v^2}{2} \right]_u^v $$

The equation for work done by a force acting on a body is given by:

\[
W = \frac{1}{2} M v^2 – \frac{1}{2} M u^2 \quad \text{…(1.13)}
\]

This result is the mathematical form of the work-energy theorem, which states that the work done by a force acting on a body is equal to the change produced in the kinetic energy of the body**.

Remember 2: If a force of varying magnitude increases the velocity of an object from $u$ to $v$, even in that case, the work done will be:

\[
W = \frac{1}{2} M v^2 – \frac{1}{2} M u^2
\]

Relation between kinetic energy and linear momentum

If a body has a mass of $ \displaystyle m$ then,

$ \displaystyle \begin{array}{l}\text{K}\text{.E}\text{.}=\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{{2m}}{{m}^{2}}{{v}^{2}}\\\Rightarrow \text{K}\text{.E}\text{.}=\dfrac{{{{p}^{2}}}}{{2m}}\end{array}$

where $ \displaystyle p$ is the linear momentum of the body i.e. $ \displaystyle p=mv$

It shows that a body cannot have kinetic energy without having linear momentum and vice-versa.

Therefore, if momentum i.e. $ \displaystyle p$ is constant the graphical representation of kinetic energy and mass is

Relation between kinetic energy and linear momentum — Image Credit: © Briligence.com

Similarly, if kinetic energy is constant, the graphical representation of kinetic momentum and mass is

and if mass is constant, the graphical representation of kinetic energy and momentum is

Example 1: A body of mass $5$ kg initially at rest is subjected to a force of $20$ N. What is the kinetic energy of the body at the end of $10$ s?

Solution:

– Given: $M = 5$ kg, $F = 20$ N, $u = 0$, $t = 10$ s.
– If acceleration produced, then
\[
a = \frac{F}{M} = \frac{20}{5} = 4 \, \text{ms}^{-2}
\]
– If $v$ is the velocity acquired by the body, then
\[
v = u + at = 0 + 4 \times 10 = 40 \, \text{ms}^{-1}
\]
– Therefore, kinetic energy of the body after $10$ s is
\[
= \frac{1}{2} M v^2 = \frac{1}{2} \times 5 \times (40)^2 = 4000 \, \text{J}
\]

Example 2: A body of mass $2$ kg initially at rest moves under the action of an applied horizontal force of $7$ N on a table with a coefficient of kinetic friction $= 0.1$. Compute the (a) work done by the applied force in $10$ s, (b) work done by friction in $10$ s, (c) work done by the net force on the body in $10$ s, (d) change in kinetic energy in $10$ s and interpret your results.

Solution:

Given: $M = 2$ kg, $u = 0$, $t = 10$ s, $\mu = 0.1$.

Applied Force, $F = 7$ N.

Force due to friction,

\[
f = \mu Mg = 0.1 \times 2 \times 9.8 = 1.96 \, \text{N}
\]

Net force under which the body moves,

\[
F_1 = F – f = 7 – 1.96 = 5.04 \, \text{N}
\]

Acceleration produced,

\[
a = \frac{F_1}{M} = \frac{5.04}{2} = 2.52 \, \text{ms}^{-2}
\]

Distance covered by the body in $10$ s,

$$ \begin{array}{l}S=ut+\dfrac{1}{2}a{{t}^{2}}\\S=0\times 10+\dfrac{1}{2}\times 2.52\times {{10}^{2}}\\S=126\text{m}\end{array}$$

(a) Work done by the applied force in $10$ s:

\[
W = F \times S = 7 \times 126 = 882 \, \text{J}
\]

(b) Work done by friction in $10$ s:

$$ \begin{array}{l}{{W}_{f}}=f\times S\\{{W}_{f}}=1.96\times 126\\{{W}_{f}}=246.96\text{J}\end{array}$

$$ \begin{array}{l}{{W}_{{net}}}={{F}_{1}}\times S=5.04\times 126\\{{W}_{{net}}}=635.04\text{J}\end{array}$$

(d) Since initial kinetic energy of the body (at rest) was zero and the final kinetic energy acquired by the body in $10$ s under the net force is $635.04$ J, the change in kinetic energy in $10$ s is $635.04$ J.

Example 3: A body of mass $1$ kg initially at rest is moved by a horizontal force of $0.5$ N on a smooth frictionless table. Calculate the work done by the force in $10$ s and show that this is equal to the change in kinetic energy of the body.

Solution:

Given: $F = 0.5$ N, $M = 1.0$ kg, $u = 0$, $t = 10$ s.

If acceleration produced,

\[
a = \frac{F}{M} = \frac{0.5}{1.0} = 0.5 \, \text{ms}^{-2}
\]

The distance traveled by the body in $10$ s is given by:

$$ \begin{array}{l}S=ut+\dfrac{1}{2}a{{t}^{2}}\\S=0\times 10+\dfrac{1}{2}\times 0.5\times {{(10)}^{2}}\\S=25\text{m}\end{array}$$

Work done by the force in $10$ s, $W = F \times S = 0.5 \times 25 = 12.5 \, \text{J}$

Let $v$ be the speed of the body after $10$ s. Then,

\[
v = u + at = 0 + 0.5 \times 10 = 5 \, \text{ms}^{-1}
\]

Change in kinetic energy:

$$ \begin{array}{l}=\dfrac{1}{2}M{{v}^{2}}-\dfrac{1}{2}M{{u}^{2}}\\=\dfrac{1}{2}\times 1\times {{0.5}^{2}}-\dfrac{1}{2}\times 1.0\times {{0}^{2}}\\=12\text{J}\end{array}$$

Thus, work done by the force in $10$ s is equal to the change in kinetic energy of the body.