Physics Lesson 5.2.2 - What are the factors affecting the Kinetic Energy of an object?

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Welcome to our Physics lesson on What are the factors affecting the Kinetic Energy of an object?, this is the second lesson of our suite of physics lessons covering the topic of Kinetic Energy, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

What are the factors affecting the Kinetic Energy of an object?

Let's try to answer this question through examples from daily life. Imagine you are a goalkeeper. Which ball is more difficult to stop when moving at the same velocity: a soccer ball or a bowling ball? Why?

Answer: A bowling ball is more difficult to stop when it is moving compared to a soccer ball as the bowling ball is heavier. From experience, we know that heavy objects are more difficult to stop when moving than lighter ones.

Another example: If you are trying to stop two identical balls, which one is more difficult to stop, the ball that is moving faster or the one that is moving slower? Why?

Answer: A fast moving ball is more difficult to stop than a slow moving identical ball.

From these examples, it is obvious that the Kinetic Energy of a moving object depends on two factors:

1. The Mass m of object, and
2. The Velocity v of the object.

The equation that relates Kinetic Energy with the abovementioned factors is

We will explain later in this topic why the velocity is at power 2 and mass is at power 1 in the Kinetic Energy formula and the reason why there is the coefficient 1/2 in the KE equation).

Example 1

A 200 g bird is flying at 6m/s. What is its Kinetic Energy in Joules?

Solution: First, let's convert grams to kilograms. Thus, 200g = 200/1000 kg = 0.200 kg.

Now, using the equation of Kinetic Energy, we obtain for the Kinetic Energy of the flying bird:

The relationship between Work and the change in Kinetic Energy of moving objects. (Work - Kinetic Energy Theorem)

When we roll a ball horizontally, eventually it will stop after travelling a certain distance (if the motion is linear we can say displacement instead of distance). This occurs because the friction does some work against the motion. Therefore, all the initial energy possessed by the ball when it starts rolling is converted into other forms of energy (heat, sound, etc) due to the friction with the ground. When the object stops moving, no energy is left in it. The change in the Kinetic Energy here is negative because

Hence, here we say friction has done a negative work.

In cases when the Kinetic Energy of an object increases due to the activation of a energy source stored inside this object (such as during explosions), the work done by these internal factors is positive as it contributes to the increase in the object's Kinetic Energy. We can write

and in the special case in which the object has initially been at rest, we can write

Thus, it is obvious the work done here by the inner factors is positive.

If we summarize all the above-mentioned findings into a single sentence (rule), we obtain the Work-Kinetic Energy Theorem, which states that

"The work done by the sum of all forces acting on an object equals the change in the kinetic energy of the object itself."

Mathematically, the Work-Kinetic Energy Theorem is written as:

Example 2

A 3 kg object slides along a horizontal surface. As a result of friction, its velocity decreases from 12m/s to 4m/s. The displacement during this process is 9.6 m. What is the average frictional force exerted between the object and the ground?

Solution. If we denote the average frictional force by < f >, we can write based on the Work - Kinetic Energy Theorem:

Substituting the values (m = 3kg; vi = 12 m/s; vf = 4 m/s; Δx = 9.6 m), we obtain

The sign minus means the frictional force f acts against motion (it is in the opposite direction of the motion).

Why is KE = ½ m × v2 and not m × v2, m × v or something else?

Let's start from the equation W = F × Δx and elaborate it. If force is not constant, we consider the average force only.

From the Newton's Second Law, we have F = m × a where m is the mass of the object and a, its acceleration. Thus, we can write

From Kinematics, it is known that one of the four equations used when an object is moving at constant acceleration (uniformly accelerated or decelerated motion), is

If we divide both sides of the last equation (eq. 4) by 2, we obtain

Now, let's substitute the left part of eq. (4) to the right part of eq. (3). Thus,

Expanding the right part of eq. (6) we obtain

Since the general mathematical expression of the Work-Kinetic Energy Theorem is written as

it is obvious that

and

In this way, we confirmed the correctness of the equation

for the Kinetic Energy of a moving object or system.

You have reached the end of Physics lesson 5.2.2 What are the factors affecting the Kinetic Energy of an object?. There are 2 lessons in this physics tutorial covering Kinetic Energy, you can access all the lessons from this tutorial below.

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4. Work, Energy and Power Practice Questions: Kinetic Energy. Test and improve your knowledge of Kinetic Energy with example questins and answers
5. Check your calculations for Work, Energy and Power questions with our excellent Work, Energy and Power calculators which contain full equations and calculations clearly displayed line by line. See the Work, Energy and Power Calculators by iCalculator™ below.
6. Continuing learning work, energy and power - read our next physics tutorial: Gravitational Potential Energy

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