Energy is the ability to do work. It can take many forms, such as heat, light, chemical energy, and electrical energy.

In recent days, we often hear about energy. Every invention and civilization is based upon acquiring and effectively utilizing energy, made possible by the unique property of our Universe - the ability to transfer and transform.

The total amount of energy is conserved; however, it can be transformed from one form to another.

One fundamental focus of physics is to investigate energy. In Physics, energy is generally defined as a scalar quantity associated with the state or condition of one or more objects.

Units of Energy

The International System of Units (SI) uses the unit of energy, Joule (J). However, energy can also be expressed in other units that do not belong to the SI system, such as calories, ergs, kilowatt-hours, kilocalories and British Thermal Units. These units must be converted to SI units in order to be used in commerce and science.

Types of Energy

In this chapter, we will mainly focus on two forms of mechanical energy:

• Kinetic
• Potential energy.

Kinetic Energy

The kinetic energy of an object with mass m and velocity v is given by the expression: $$KE = \frac{1}{2}mv^2$$ The greater the kinetic energy, the faster the object moves.

$K=\frac{1}{2}m{v}^2$

Kinetic energy: Kinetic energy is the energy of motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

A 3kg block moving past you at 2.0m/s has a kinetic energy of 6.0J.

Potential Energy

It is associated with the configuration (arrangement) of the system of objects that exert forces on one another.

Example of Potential Energy:

The energy stored in a stretched or compressed spring.

Potential Energy Transforms into Kinetic Energy

The increase in kinetic energy of the ball when thrown from a height h above the surface of the earth is accounted for by the (gravitational potential energy). This energy is associated with the configuration of two objects (ball and earth) being attracted by the gravitational force.

Work and Potential Energy

At the maximum height of the ball, its velocity (kinetic energy) has decreased to zero. Therefore, the energy it possesses at this point is gravitational potential energy, which is the result of a negative work done by the gravitational force as the ball was thrown up.

The work done by the gravitational force is stored as potential energy. Therefore, for any case, the change in potential energy is equal to the negative of the work done on the system.

$\mathrm{\Delta }U=-W=-F\cdot \mathrm{\Delta }x$

$F\ast \Delta x=Constant$

If the force is a conservative force, then the change in potential energy is given by:

${\displaystyle \Delta U={\int }_{r1}^{r2}f\left(r\right)dr}$

The potential energy of a conservative force can be found by:

\begin{array}{l}F(r) = -\frac{dU}{dx} = 0\end{array}

Potential Energy and Equilibrium

The object is said to be in equilibrium if the net force acting on it is equal to zero. $$F(r) = -\frac{dU}{dx} = 0$$

A body is in stable equilibrium if, upon slight displacement, a restoring force acts on it and brings it back to its normal position.

\(\frac{d^2U}{dx^2} > 0\)

A body is in unstable equilibrium if, upon slight displacement, the restoring force does not cause the object to return to its original position.

\(\frac{d^{2}U}{dx^{2}} < 0\)

Neutral equilibrium: We can move the particle slightly away from such a point and it will still remain in equilibrium (i.e., it will neither attempt to return to its initial state nor will it continue to move) $$\frac{d^{2}U}{dx^{2}} = 0$$

Problem:

The potential energy of a conservative system is given by U = ax² - bx, where a and b are positive constants.

| This problem involves the use of potential energy and conservative force relation |

Find the equilibrium position and discuss whether the equilibrium is stable, unstable, or neutral.

\begin{array}{l}F = \frac{dU}{dx}\end{array} in a conservation field

$\therefore F=-\frac{d}{dx}(a{x}^2-bx)=2ax+b$

For equilibrium, $$F = 0$$ or $$b - 2ax = 0$$

$\therefore x=\frac{b}{2a}$

We can see from the given equation that $\frac{d^{2}U}{d{x}^2}$ is positive and equal to ( 2a ), meaning that ( U ) is a minimum.

Therefore, $x=\frac{b}{2a}$ is the stable equilibrium position.

Gravitational Potential Energy

The work done by the gravitational force for displacing an object from height 0 to h is given by:

\begin{array}{l}W = \int_{0}^{h}mg ;dy\end{array}

$\mathrm{\Delta }U=-mg\cdot h$

$\mathrm{\Delta }U=mgh$

Check Out: Gravitation

Elastic Potential Energy

The change in elastic potential energy for a spring-mass system with a spring of spring constant k and a block of mass m displaced through a distance x from its equilibrium position (x=0) is given by:

$$\Delta U_e = -kx^2/2$$

$\mathrm{\Delta }U=\frac{-k{x}^2}{2}{|}_0^x$

$\mathrm{\Delta }U=k\left[\frac{x^2}{2}\right]$

Problem:

A block of mass m is attached to two unstretched springs of constant k₁ and k₂ as shown in the figure. The block is displaced towards the right through a distance of x and released. Find the speed of the block as it passes through a distance x/4 from its mean position.

Solution:

The speed of the block as it passes through a distance x/4 from its mean position can be found by solving the differential equation of motion of the block, given the mass m, the constants k₁ and k₂, and the displacement x.

Using energy conservation

\begin{array}{l}K_1x^2 + K_2x^2 = mv^2\end{array}

\begin{array}{l}\frac{1}{2}k_1(x/4)^2 + \frac{1}{2}k_2(x/4)^2 = \frac{1}{2}mv^2\end{array}

\begin{array}{l}v= \frac{x}{4m}\sqrt{(k_1+k_2)}\end{array}

Work-Energy Theorem

This theorem states that the total work done by all the forces acting on a particle or body is equal to the change in its kinetic energy.

The example given is of a block of mass m kept on a rough horizontal surface, acted upon by a constant force F parallel to the surface, and displaced through x. The initial and final velocities are v₀ and v, respectively.

Now, we apply Newton’s law:

$$F = ma = m\left ( \frac{v^{2}-v_{o}^{2}}{2x} \right )$$

Finally, the work is done.

\begin{array}{l}W = Fx = \frac{1}{2}mv^{2}-\frac{1}{2}mv_{o}^{2}\end{array}

Definition: The Kinetic energy (K) of an object is the energy it possesses due to its motion.

\begin{array}{l}K = \frac{1}{2}mv^{2}\end{array}

The work done on an object is equal to the change in the object’s [energy].

\begin{array}{l}W = K - K_o = \Delta K\end{array}

Problem:

1. The displacement of a body in a meter is a function of time according to x = 2t⁴+5. The mass of the body is 2kg. What is the increase in its kinetic energy one second after the start of motion?

Solution:

1. The increase in kinetic energy of the body with a mass of 2kg one second after the start of motion, according to the equation x = 2t⁴+5, is 80J.

• a 8J
• b 16J
• c 32J
• d 64J

Solution : d.

X = 2t⁴ + 5

\begin{array}{l}\Rightarrow \frac{dx}{dt} = 8t^{3}\end{array}

$$a = \frac{dv}{dt} = 24t^{2}$$

$$F = ma = 48t^2$$

\begin{array}{l}dW = Fdx = 384t^{5}dt\end{array}

The work done by the applied force results in an increase in the kinetic energy.

$\mathrm{\Delta }KE=\frac{48\times 8}{6}=64J$

What is the change in kinetic energy of the 2kg block when a force shown in the F - x graph is applied to it horizontally on a frictionless horizontal surface?

a) 15J

20J

25J

30J

Solution :

Work done = Area under F-x graph

W = (1/2) × (10 - 2) × 5 = 20J

Work done = Change in kinetic energy = 20J

Law of Conservation of Energy

Conservation of energy means the total energy within an isolated system remains constant, accounting for all forms of energy together. This includes the two forms of mechanical energy, namely kinetic energy (K) and potential energy (U).

The total mechanical energy of a system remains constant if only conservative forces act on it, i.e. (K+U = constant).

Example

If the pendulum is given a velocity from its equilibrium position, it will undergo a simple harmonic motion wherein energy is conserved at all points of its trajectory.

As the bob is raised from the equilibrium point, its position above the surface of the earth increases, resulting in an increase in potential energy. This is accompanied by a decrease in kinetic energy as the bob slows down while raising.

At all points, total mechanical energy is conserved. When kinetic energy is maximum (at equilibrium position), potential energy is minimum and when potential energy is maximum (extreme positions), kinetic energy is minimum.

Problem:

The height from which the mass was dropped can be calculated using the equation $h=\frac{KE}{\frac{1}{2}m{g}^2}$, where KE is the kinetic energy, m is the mass, and g is the acceleration due to gravity. In this case, the height is $h=\frac{400J}{\frac{1}{2}(2.0kg)(9.8m/s^2)}=81.63m$.

By conservation of mechanical energy,

U_f + K_f = U_i + K_i

0 + K_f = mgh + 0

\begin{array}{l} \Rightarrow h = \frac{K_f}{mg} = \frac{400}{2 \times 9.8} = 20.4 \text{m} \end{array}

Question 2

If a 10 kg block is released from the top of the incline and is brought to rest momentarily after compressing the spring 2.0m, then the mass slides through a distance of 1.0m before coming to rest.

K = 100 N/m

The mass describes a distance of x before reaching the spring. Then the total distance described is (x + 2) before coming to rest.

Work done by gravitational pull = $10g(x+2)\sin 30°$

5g(x + 2)

-Work done by the elastic force of the spring = -(1/2)$$kx^2$$

-100 x 22 = -2200

Net work done on the body = $5g\times (x+2)-200$

The Work-Energy Theorem: $\mathrm{\Delta }K=0$

5gx + 10g - 200 = 0

x + 2 = 20

x = 2

x = 2 along the inclined surface

Frequently Asked Questions about Energy

What is the unit of energy?

Joule is the unit of energy.

What are the two main categories of energy?

Potential Energy

Kinetic Energy

The work-energy theorem states that the work done on an object is equal to the change in the kinetic energy of the object.

According to the Work-Energy Theorem, the net work done on an object by a force is equal to the change in its kinetic energy.