Law of Conservation of Energy - A Complete Guide

The Law of Conservation of Energy governs everything around us, from the way our bodies use food to produce energy, to how machines operate and even how stars shine in the universe. By exploring the Law of Conservation of Energy which is a fundamental principle in Physics, we can see how energy transfers and transformations impact our everyday lives, ensuring that nothing is ever truly lost, just changed. Let's read this entire article to gain an in-depth understanding of law of conservation of energy

Law of Conservation of Energy

According to law of conservation of energy "Energy can neither be created nor destroyed but it can only be transformed from one form to another." The total energy in an isolated system is constant.

Mathematically it is expressed as,

$E_{\text {total }}=$ constant where, $E_{\text {total }}$ represents the all forms of energy.

Derivation of Law of Conservation of Energy

At the top of the tree, the fruit is at a height $H$ from the ground, and its total energy is purely potential energy (P.E.):

$
E=m g H ..........(1)
$

As the fruit falls freely under gravity, its potential energy decreases, and kinetic energy increases. At point $B$, which is at a height $X$ from the ground, the fruit has both potential energy and kinetic energy:

$
E=\text { K.E. + P.E. }
$
The potential energy at this point is:

$
\text { P.E. }=m g X.........(2)
$
To find the kinetic energy (K.E.), we use the third equation of motion.

From the third equation of motion:

$
v^2=2 g(H-X)
$
Here, $v$ is the velocity of the fruit at height $X$. The kinetic energy at this point is given by:

$
\text { K.E. }=\frac{1}{2} m v^2
$
Substituting $v^2=2 g(H-X)$ into the equation for K.E.:

$
\begin{gathered}
\mathrm{K} . \mathrm{E} .=\frac{1}{2} m \cdot 2 g(H-X) \\
\mathrm{K} . \mathrm{E} .=m g(H-X)........(3)
\end{gathered}
$

Now, combining the expressions for K.E. (Equation 3) and P.E. (Equation 2):

$
\begin{gathered}
E=\mathrm{K} . \mathrm{E} .+\mathrm{P} . \mathrm{E} . \\
E=m g(H-X)+m g X
\end{gathered}
$
Simplifying:

$
\begin{gathered}
E=m g(H-X+X) \\
E=m g H...........(4)
\end{gathered}
$

Also read -

• NCERT Solutions for All Subjects
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Conservation of Mechanical Energy

Mechanical energy is the sum of potential energy and kinetic energy.

According to Conservation of Mechanical Energy, If only conservative forces act on a system, the total mechanical energy remains constant.

By work-energy theorem, we have
$
W=k_f-k_i \text { or } \triangle K=\int_{r_i}^{r_f} \vec{f} \cdot d \vec{s}
$

The change in potential energy in a conservative field is
$
U_i-U_f=\int_{r_i}^{r_f} \vec{f} \cdot \overrightarrow{d s}
$

Or,
$
-\triangle U=\int_{r_i}^{r_f} \vec{f} \cdot \overrightarrow{d s}
$

From equation (1) and (2)
We get, $\Delta K=-\Delta U$
$
\Delta K+\Delta U=0
$

Means, $K+U=E$ (constant)
$\mathrm{Or}, \mathrm{E}$ is constant in a conservative field i.e.; if the kinetic energy of the body increases its potential energy will decrease by an equal amount and vice versa.

Related Topics

• Elastic And Inelastic Collision
• Basics Of Energy And Its Various Forms
• Moment of inertia
• Collision

Examples of Law of Conservation of Energy

1. Potential energy is converted into kinetic energy in a free-falling object, but the total energy remains constant.
2. When fuel burns in a Car engine then chemical energy in the fuel converts to thermal energy during combustion.
3. In an electric bulb, electrical energy is converted into light energy and heat energy.
4. When an archer pulls back a bowstring, mechanical work is stored as elastic potential energy in the stretched bow.
5. When a pendulum swings, its energy alternates between potential energy at the highest points of the swing and kinetic energy at the lowest point.

Solved Example Based on Law of Conservation of Energy

To further comprehend the above concept, consider the solved example provided below.

Example 1: Two hydrogen atoms are in an excited state with electrons residing in n = 2. The first one is moving towards the left and emits a photon of energy $E_1$ towards the right. The second one is moving towards the right with the same speed and emits a photon of energy $E_2$ towards the right. Taking the recoil of the nucleus into account during the emission process
1)$
E_1>E_2
$
2)$
E_1<E_2
$
3)$
E_1=E_2
$
4) none

Solution:

If only conservative forces act on a system, total mechanical energy remains constant -
$
\begin{aligned}
& K+U=E \text { (constant }) \\
& \Delta K+\Delta U=0 \\
& \Delta K=-\Delta U
\end{aligned}
$

In the first case, K.E. of $\mathrm{H}$-atom increases due to recoil whereas in the second case, K.E. decreases due to recoil but
$
\begin{aligned}
& E_1+K E_1=E_2+K E_2 \\
\therefore & E_2>E_1
\end{aligned}
$

Example 2: A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is the acceleration due to gravity, the work required to pull the hanging part onto the table is

1) MgL
2) $\frac{M g L}{3}$
3) $\frac{M g L}{9}$
4) $\frac{M g L}{18}$

Solution:

Work done to raise the centre of mass of the chain on the table is given by

$W=\frac{M g L}{2 n^2}$

As 1/3 part of the chain is hanging from the edge of the table. So by substituting n = 3 in standard expression

We get : $W=\frac{M g L}{2 n^2}=\frac{M g L}{2(3)^2}=\frac{M g L}{18}$

Hence, the answer is the option 4.

The velocity of the Chain While Leaving the Table

Taking the surface of the table as a reference level (zero potential energy)

The potential energy of the chain when $\left(\frac{1}{n}\right)^{t h}$ length hanging from the edge $=-\frac{M g L}{2 n^2}$
The potential energy of the chain when it leaves the table $=-\frac{M g L}{2}$ (here $\left.\mathrm{n}=1\right)$ By the law of conservation of mechanical energy,
$
\begin{aligned}
& K \cdot E_{\cdot}+U_i=K \cdot E_{\cdot f}+U_f \\
\Rightarrow & 0-\frac{M g L}{2 n^2}=\frac{1}{2} M v^2-\frac{M g L}{2} \\
\Rightarrow & \frac{1}{2} M v^2=\frac{M g L}{2}-\frac{M g L}{2 n^2} \\
\Rightarrow & v=\sqrt{g L\left[1-\frac{1}{n^2}\right]}
\end{aligned}
$

Example 3: A man places a chain (of mass ‘m’ and length ‘l’) on a table slowly. Initially, the lower end of the chain just touches the table. The man drops the chain when half of the chain is in a vertical position. Then, the work done by the man in this process is :

1) $-m g \frac{l}{2}$
2) $-\frac{m g l}{4}$
3) $-\frac{3 m g l}{8}$
4) $-\frac{m g l}{8}$

Solution:

Potential Energy stored when a particle is displaced against gravity -

\begin{aligned}
& U=-\int f d x=-\int(m g) d x \cos 180^{\circ} \\
& \text { - wherein } \\
& m=\text { mass of body } \\
& g=\text { acceleration due to gravity } \\
& d x=\text { small displacement }
\end{aligned}

The work done by man is negative of the magnitude of the decrease in potential energy of the chain

$\begin{aligned} & \Delta U=m g \frac{L}{2}-\frac{m}{2} g \frac{L}{4}=3 m g \frac{L}{8} \\ & \therefore W=-\frac{3 m g l}{8}\end{aligned}$

Example 4: A ball of mass 4kg moving with a velocity of 10 ms^-1, collides with a spring of length 8m and force constant 100 Nm^-1. The length of the compressed spring is x m. The value of x, to the nearest integer is _______.

1) 6

2) 2

3) 8

4) 10

Solution:

Let the spring be compressed by y.

Applying the energy conservation principle,

$\begin{aligned} & \Rightarrow \frac{1}{2} m v^2=\frac{1}{2} k y^2 \\ & \Rightarrow y=\sqrt{\frac{m}{k}} \cdot v \\ & \Rightarrow y=\sqrt{\frac{4}{100}} \times 10 \\ & \Rightarrow y=2 m\end{aligned}$

Therefore, the final length of the spring =8−2 = 6m

Hence, the answer is (6).

Example 5: Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural length L and spring constant K. A third block C of mass m moves with a speed v along the line joining A and B collides with A. The maximum compression in the spring is

1) $\sqrt{\frac{m}{2 K}}$
2) $v \sqrt{\frac{m}{2 K}}$
3) $\sqrt{\frac{m v}{K}}$
4) $\sqrt{\frac{m v}{2 K}}$

Solution:

If only conservative forces act on a system, total mechanical energy remains constant -

$
\begin{aligned}
& K+U=E(\text { constant }) \\
& \qquad V=\frac{x^4}{4}-\frac{x^2}{2} .
\end{aligned}
$

First, we need to calculate the minimum potential energy.
When potential energy is minimum
$
\begin{aligned}
& \frac{d V}{d x}=0 \Rightarrow x^3-x=0 \\
& \therefore x=0 \text { or } x=1 \\
& \mathrm{~V}(\mathrm{x}=0)=0 \\
& V(x=1)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4} J \\
& \text { So minimum potential energy }=\frac{-1}{4} J
\end{aligned}
$

So minimum potential energy
K.E. + P.E. $=$ Total mechanical energy.
$K . E .\left.\right|_{\max }+$ P.E. $\left.\right|_{\min }=$ Total mechanical energy
$
\begin{aligned}
& K \cdot E_{\cdot \max }-\frac{1}{4}=2 J \\
& K \cdot E_{\cdot \max }=\frac{9}{4} J
\end{aligned}
$

$\begin{aligned} \Rightarrow & \frac{1}{2} m V_{\max }^2=\frac{9}{4} \text { or } V_{\max }^2=\frac{9}{2} \\ & V_{\max }=\frac{3}{\sqrt{2}}\end{aligned}$

Hence, the answer is option (2).