 # Energy - Kinetic and Potential Energy

## What is Energy?

In recent days, we often hear about energy. Every invention, civilization is based upon acquiring and effectively using energy. This is possible by the unique property of our Universe energy that it can be transformed and transferred, but the total amount is always the same (conserved). One fundamental focus of physics is to investigate energy.

Energy can be generally defined as a scalar quantity associated with the state or condition of one or more objects. In this chapter, we will mainly focus on two forms of mechanical energy: Kinetic and Potential energy.

### Kinetic energy

It is associated with the state of motion of an object. The greater the kinetic energy, the faster it moves. For an object of mass m and velocity v, it is given by the expression,

$K = \frac{1}{2}mv^{2}$

Unit of energy (all types) is joule (J)

Example:

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: When you throw a ball down from a height h above the surface of the earth, the increase in kinetic energy of the ball is accounted by defining gravitational potential energy. This energy is associated with the state of separation (configuration) between two objects (ball and earth) attracted by gravitational force.

### Work and Potential energy

In the above example, the work done by the gravitational force is negative if the ball is thrown up. As the ball rises, its velocity (kinetic energy) decreases and reaches zero when it attains the maximum height. Now, we can tell that the energy the ball possess at the maximum height is gravitational potential energy. The work done by the gravitational force is stored as the potential energy. Hence, for any case, the change in potential energy is defined as being equal to the negative of the work done on the system.

$\Delta U = -W = -F.\Delta x$

Where F is the constant force and $\Delta x$ is the displacement due to the force.

More generally, if the force is a conservative force, then the change in potential energy is given by,

$\Delta U = \int_{r_1}^{r_2}F(r)dr$

Conservative force from potential energy can be found by,

$F(r) = -\frac{dU}{dx} = 0$

### Potential energy and equilibrium

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

Stable equilibrium: A body is in stable equilibrium if it comes back to its normal position on slight displacement. For slight displacement, the restoring force should act on it.

$\frac{d^{2}U}{dx^{2}}> 0$

Unstable equilibrium: A body is in unstable equilibrium if it does come back to its normal position on slight displacement. For slight displacement, the restoring force should take away the object.

$\frac{d^{2}U}{dx^{2}}< 0$

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 is an application of potential energy and conservative force relation

Find the equilibrium , position and discuss whether the equilibrium is stable, unstable or neutral. In a conservation field $F = \frac{dU}{dx}$

∴ $F = -\frac{d}{dx}(ax^{2}-bx) = b-2ax$

For equilibrium F = 0 or b – 2ax = 0

∴ $x = \frac{b}{2a}$

From the given equation we can see that $\frac{d^{2U}}{dx^{2}} = 2a$ (positive)., U is minimum

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

## Gravitational Potential Energy

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

$W = \int_{r_1}^{r_2}F.dr = \int_{0}^{h}mg \;dy$ $\Delta U = -\int_{0}^{h}-mg\;dy$ $\Delta U = mgh$ ### Elastic Potential Energy

Let’s consider 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). As the block moves from x=0 to x, the spring force, Fs=-kx does work on the block. The corresponding change in elastic potential energy is given by,

$\Delta U = -\int_{0}^{x}-kx\;dx$ $\Delta U = k[\frac{x^{2}}{2}]$

Problem:

A block of mass m is attached to two unstretched springs of spring constant k₁ and k₂ as shown in figure . The block is displaced towards 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.

Applying conservation energy $\frac{1}{2}K_1x^{2}+\frac{1}{2}K_2x^{2} = \frac{1}{2}mv^{2}+\frac{1}{2}k_1(x/4)^{2}+\frac{1}{2}k_2(x/4)^{2}$ $v= \frac{x}{4m}\sqrt{15(k_1+k_2)}$

## Work- Energy theorem

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

Let us take an example in which a block of mass m kept on a rough horizontal surface is acted upon by a constant force F parallel to the surface and it is displaced through x. The initial velocity and the final velocity are respectively, v0 and v. Now, we apply Newton’s law:

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

Finally, the work done is

$W = Fx = m\left ( \frac{v^{2}-v_{o}^{2}}{2x} \right )x =\frac{1}{2}mv^{2}-\frac{1}{2}mv_{o}^{2}$

We now define the Kinetic energy(K) of the object:

$K = \frac{1}{2}mv^{2}$

By this definition, the work done on the object is simply equal to the change in the object’s

$W = K-K_o = \Delta K$

Problem:

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

1. 8J b)16J

c) 32J d)64J

D)

X = 2t4 +5

$\Rightarrow v = \frac{dx}{dt} = 8t^{3}$ $a = \frac{dv}{dt} = 24t^{2}$ $\Rightarrow F = ma = 48t^{2}$ $dW = Fdx = 48t^{2}\times8t^{3}dt$

Increase in the kinetic energy results from the work done by the applied force

$\Delta KE = \int_{0}^{1}48\times8t^{5}dt = \frac{48\times8}{6}t^{6} = \frac{48\times8}{6} = 64J$

A 2kg block is placed on a frictionless horizontal surface. A force shown in the F – x graph is applied to the block horizontally. The change in kinetic energy is a) 15J b) 20J

c) 25J d) 30J

b)

Work done =Area under F-x graph

W = 1/2 x(10-2)x5 = 20J

Work done = change in kinetic energy = 20J

### Law of conservation of energy

Conservation of energy means conservation of all forms of energy together. Accounting all forms of energy within an isolated system, the total energy remains constant. The mechanical energy accounts for only two forms of energy, namely kinetic energy, K and potential energy, U.

If only conservative forces act on a system, then total mechanical energy of the system remains constant. i.e. $K+U = constant$ Example

Consider the case of a simple pendulum of massive bob hanging at the end of a massless rod that is pivoted at a fixed point. If the pendulum is given a velocity from its equilibrium position, it executes a simple harmonic motion where energy is conserved at all points.

As the bob is raised from the equilibrium point, its position above the surface of the earth is increased and hence the potential energy is increased. It is accounted for the decrease in kinetic energy as the bob slows down while raising.  When kinetic energy is maximum (at equillibrium position), potential energy is minimum. When potential energy is maximum (extreme positions), kinetic energy is minimum. Thus at all points, total mechanical energy is conserved.

Problem:

Just before striking the ground, a 2.0kg mass has 400J of kinetic energy. If friction can be ignored, from what height was it dropped?(g = 9.8m/s2)

By conservation of mechanical energy,

Uf+Kf = Ui+Ki

or,0+Kf = mgh+0

$\Rightarrow h=\frac{K_f}{mg} = \frac{400}{2\times9.8} = 20.4m$ An idea massless spring ‘S’ can be compressed 1.0m by a 100N force. It is placed as shown at the bottom of a frictionless inclined plane which makes an angle of $\theta = 30^{o}$ with the horizontal. A 10 kg block is released from the top of the incline and is brought to rest momentarily after compressing the spring 2.0m. Through what distance does the mass slide before coming to rest ?

Spring constant = 100 N\m

Conserving energy,

$mg(x+0) \sin 30^{o} = \frac{1}{2}K.(2)^{2}$ $10\times10(x+2)\times \frac{1}{2} = \frac{1}{2}\times100$ $x+2 = 4$ x = 2 along the inclined surface