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# What is kinetic energy?

Learn what kinetic energy means and how it relates to work.

# What is kinetic energy?

Kinetic energy is the energy an object has because of its motion.
If we want to accelerate an object, then we must apply a force. Applying a force requires us to do work. After work has been done, energy has been transferred to the object, and the object will be moving with a new constant speed. The energy transferred is known as kinetic energy, and it depends on the mass and speed achieved.
Kinetic energy can be transferred between objects and transformed into other kinds of energy. For example, a flying squirrel might collide with a stationary chipmunk. Following the collision, some of the initial kinetic energy of the squirrel might have been transferred into the chipmunk or transformed to some other form of energy.

# How can we calculate kinetic energy?

To calculate kinetic energy, we follow the reasoning outlined above and begin by finding the work done, W, by a force, F, in a simple example. Consider a box of mass m being pushed through a distance d along a surface by a force parallel to that surface. As we learned earlier
\begin{aligned} W &= F \cdot d \\ &= m · a · d\end{aligned}
If we recall our kinematic equations of motion, we know that we can substitute the acceleration if we know the initial and final velocity—v, start subscript, i, end subscript and v, start subscript, f, end subscript—as well as the distance.
\begin{aligned} W &= m\cdot d\cdot \frac{v_\mathrm{f}^2-v_\mathrm{i}^2}{2d} \\ &= m\cdot \frac{v_\mathrm{f}^2-v_\mathrm{i}^2}{2} \\ &= \frac{1}{2}\cdot m \cdot v_\mathrm{f}^2 - \frac{1}{2}\cdot m \cdot v_\mathrm{i}^2 \end{aligned}
So, when a net amount of work is done on an object, the quantity start fraction, 1, divided by, 2, end fraction, m, v, squared—which we call kinetic energy K—changes.
start text, K, i, n, e, t, i, c, space, E, n, e, r, g, y, colon, space, end text, K, equals, start fraction, 1, divided by, 2, end fraction, dot, m, dot, v, squared
Alternatively, one can say that the change in kinetic energy is equal to the net work done on an object or system.
W, start subscript, n, e, t, end subscript, equals, delta, K
This result is known as the work-energy theorem and applies quite generally, even with forces that vary in direction and magnitude. It is important in the study of conservation of energy and conservative forces.

# What is interesting about kinetic energy?

There are a couple of interesting things about kinetic energy that we can see from the equation.
• Kinetic energy depends on the velocity of the object squared. This means that when the velocity of an object doubles, its kinetic energy quadruples. A car traveling at 60 mph has four times the kinetic energy of an identical car traveling at 30 mph, and hence the potential for four times more death and destruction in the event of a crash.
• Kinetic energy must always be either zero or a positive value. While velocity can have a positive or negative value, velocity squared is always positive.
• Kinetic energy is not a vector. So a tennis ball thrown to the right with a velocity of 5 m/s, has the exact same kinetic energy as a tennis ball thrown down with a velocity of 5 m/s.
Exercise 1a: Being in the wrong place when an African elephant—mass = 6000 kg, velocity = 10 m/s—is charging can really ruin your day. How fast would a 1 kg cannon ball travel if it had the same kinetic energy as the elephant?
Exercise 1b: How would you expect the damage done to a brick wall to differ in the event of separate collisions with the elephant and cannonball?
Exercise 2: Hydrazine rocket propellant has an energy density E, start subscript, d, end subscript of 1, point, 6, start fraction, start text, M, J, end text, divided by, start text, k, g, end text, end fraction. Suppose a 100 kg (m, start subscript, r, end subscript) rocket is loaded with 1000 kg (m, start subscript, p, end subscript) of hydrazine. What velocity could it achieve? To keep things simple, let’s assume that the propellant is burned up very quickly and that the rocket is not subject to any external forces.

## Want to join the conversation?

• in the exercise 2 , how is the total chemical potential energy stored in the propellant is (Energy density) x (mass of hydrazine)??
• By definition, "energy density" tells you energy per mass for the propellant. In other words, energy density is equal to the amount of energy contained in each kilogram of propellant. Multiplying by the number of kilograms of propellant will thus give you the total energy.
• I don't understand what net work means. It says in the article that kinetic energy can be equal to the net work done on a system. What does net work mean?
• Net-work just means the total work done by the system. And the value of net work = change in kinetic energy.
• prove that K.E = 1/2 mv2?
• You are asking a question, which is answered very well in the current topic :( However, here is somewhat a different proof based on the same idea, although I tried to increase the understanding. Hope this helps.
To prove that Kinetic Energy is ½ mv^2

We already know that Energy w = f x s that is force x displacement.
So,
ΔK = W = FΔs = maΔs

From the third equation of motion that is v^2 - u^2 = 2aΔs (-1)
where
v = final velocity, u = initial velocity, a = acceleration, and Δs denotes change in displacement

ΔK = maΔs, so from equation (1)
 ΔK = m(v^2 - u^2) / 2=> ½ mv^2 - ½ mu^2

but If kinetic energy is the energy of motion then, naturally, the kinetic energy of an object at rest should be zero. Therefore, we don't need the second term and an object's kinetic energy is just
½ mv^2
• Why is kinetic energy related to the square of the velocity, as opposed to momentum which varies linearly with velocity? How do I know when I'm dealing with a momentum problem or a kinetic energy problem?
• The simplest derivation is this..... Work W=Fx sort of definition.
Then putting F=ma and x=(1/2)at^2 in above gives
W=(1/2)m(a^2)(t^2)
=(1/2)m v^2
This is called the kinetic energy, KE.
If you realise it doesn't matter how one arrives at the given speed v then this result must be quite general.
• if there is the another way to say kinetic energy
• if the propellant is contained by rocket then the K.E of rocket shouldn't be => K.E= 0.5(mr+mp)(v)^2
i know it's incorrect but can't figure out .... please clarify , its confusing me
plus , how can the K.E of rocket equal to the CHEM P.E of hydrazine?
• The propellant is used up as the rocket gains speed, right? When the propellant is all gone, the rocket's KE has to be equal to whatever energy was produced by the propellant as it burned (actually the propellant leaves as gas and that gas has KE, too, so it' really the sum of that KE plus the KE of the rocket that has to add up)
• For exercise 2, I want to know the units of factors used in the last step to calculate the V value. When I calculated the velocity with Ed = 1.6MJ/kg, Mp = 1000kg, and Mr = 100kg, I got 5.657. It looks like 1000 was multiplied to the value, but why? Is it because of the MJ to J? But isn't 1megajoule = 1000000J?
• Yes, you should convert MJ to J. The reason that your answer must me multiplied by 1000 since sqrt(10^6)=10^3=1000