# Calculating kinetic energy of an object

Energy is available to us in many forms: electromagnetic, thermal, gravitational, chemical, and so on. By the principle of conservation of energy, any of these forms of energy may be converted into the energy of motion of an object, the kinetic energy. Recall that the kinetic energy in one dimension was defined as:

\[K = \frac12 mv^2.\]

This formula holds for translational motion of a mass \(m\) at velocity \(v\). For more complicated motion such as rotation, there are additional contributions to the energy of motion (e.g. the rotational kinetic energy). In this article, some cases of computing the kinetic energy of an object given that the sum of all forms of energy is conserved.

## Work and Energy in One Dimension

Kinetic energyis the energy possessed by an object in motion. It is a scalar quantity, meaning it only possesses a magnitude. It is typically measured in Joules (J).

Energy is closely related to the concept of work, since doing work on objects can increase or decrease their energy. The work done by a constant force \(F\) acting on an object in one dimension is the product of the force times the distance through which the object moves:

\[W = F\cdot d.\]

If the direction of the force or magnitude of the force changes over time, this formula can be generalized doing calculus.

According to the **work-energy theorem**, the work done on an object is entirely converted into the kinetic energy of the object. This is a consequence of the conservation of energy; in order to do work on an object, some amount of energy must be expended. In order to conserve energy, the expended energy is made up in the additional kinetic energy of the object.

For some forces, like friction, the work done by the force actually removes kinetic energy from an object. In this case, one says that the force does *negative* work on the object.

If person A has walked \(2\) km, and person B has walked \(5\) km, who has spent more energy?

Solution:

It is common sense that person B has spent more energy. The formal reason for this is that in walking, people do work against gravity in lifting their feet (or work to overcome friction by dragging their feet). So when walking, people constantly exert force in the direction of their movement to keep them moving, i.e. they do work. People who travel further will do more work since they will be exerting this force over a larger amount of distance. This distance may be less than the actual distance walked, but it is a safe assumption that it is roughly proportional to distance traveled. Therefore, person B has spent more energy.

Person A applies \(50\)N of force on an object and person B \(100\)N of force on the same object for some fixed amount of time \(t\). Who spends more energy?

Solution:

If neither are able to move the object, than neither person spends more energy. If both people are able to move the object, however, then person B spends more energy. Since it is the same object, both people are able to give a momentum \(p = Ft\) to the object if \(F\) is the force they apply. Kinetic energy in one dimension is equal to \(p^2 / (2m)\) where \(m\) is the mass of the object. Whoever exerts the larger force for a fixed amount of time therefore imparts more momentum and thus more energy to the object.

## Example Problems

## A force of \(5 \text{ N}\) was applied to an object (which was initially at rest) over a distance of \(5 \text{ m}\). Calculate the kinetic energy of the object after the force was applied.

Solution:

The work done on the object by the force is:

\[W = F\cdot d = (5)(5) \text{ J} = 25 \text{ J}\]

Therefore, the object has a kinetic energy of \(25 \text {J}\) after the force was applied.

## An object with a mass of \(10 \text{ kg}\) is moving with a velocity of \(5 \text{ m}/\text{s}\). Calculate its kinetic energy.

Solution:

\[E_k = \frac{1}{2}mv^2 \\ E_k = \frac{1}{2}(10)(5)^2 \\ E_k = 125 \text{ J}\]

The object has a kinetic energy of \(125 \text {J}\).

Person A pushes a \(2 \text{ kg}\) mass constantly with some force. The mass moves on a rough surface with coefficent of friction \(\mu = 0.5\). The mass moves a total of \(1 \text{ m}\) at constant velocity of \(.2 \text{ m}/\text{s}\) while person A is pushing.

In a second experiment, a ball of mass \(.5 \text{ kg}\) falls from a height of \(2 \text{ m}\). Person B catches it after falling \(2 \text{ m}\), bringing it to rest.

Who does more work in total?

A ball of mass \(m\) and velocity \(v\) collides inelastically with an identical ball, coming to rest and imparting \(50\)% of its energy to the second ball. This ball then collides inelastically with a third identical ball, coming to rest and imparting \(50\)% of its energy to the third ball. What is the final velocity of the third ball?

## Simple Derivation of Work-Energy Theorem

Below is a simple proof of the work-energy theorem in one dimension assuming a constant force.

The work \(W = F\cdot d\) exerted on an object is entirely converted into kinetic energy \(K = \frac12 mv^2\).

Consider a body of mass \(m\) initially at rest on a smooth floor. Let a force \(F\) act upon it so that the body attains certain velocity \(v\), and suppose it travels a distance \(d\) in the process.

On one hand, the work done in accelerating the object up to velocity \(v\) is \(W=Fd\). On the other hand, since the body has accelerated (assume some constant acceleration \(a\) and that the force is in the direction of motion), the force can be written in terms of the acceleration as \(F = ma \implies W = mad\).

From the equations of one dimensional kinematics, the object's velocity obeys:

\[v^2=2ad\implies d= \dfrac{v^2}{2a}\]

Substituting in this expression for the total distance traveled into the expression for the work done on the object, one finds:

\[W=ma\times \dfrac{v^2}{2a} \\ = \dfrac12 mv^2\]

which is just the change in kinetic energy of the object.

**Cite as:**Calculating kinetic energy of an object.

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