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.

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.

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.

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

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)

Can you say that kinetic energy is the source of sound energy? Can you say that kinetic energy and sound energy exist together?

Sound energy is a type of 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 ```

If you see in first paragraph,how will object move with a constant speed if I apply energy only once? It will stop after a certain interval of time

If it stops, that means someone or something else was "applying energy". In physics we call that "doing work". Friction does work to stop the object after you did work to get it moving. If you did it in outer space the object wouldn't stop

Main content

What is kinetic energy?

Google Classroom

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}

[Huh? I'm lost already.]

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.

[What kinematic formula is this?]

\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?

[Show solution.]

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?

[Show solution.]

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.

[Show solution.]

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Bhoovesh Waran
Posted 7 years ago. Direct link to Bhoovesh Waran's post “in the exercise 2 , how i...”
in the exercise 2 , how is the total chemical potential energy stored in the propellant is (Energy density) x (mass of hydrazine)??
Button navigates to signup pageComment on Bhoovesh Waran's post “in the exercise 2 , how i...”
(9 votes)
Answer
- vinnv226
  Posted 7 years ago. Direct link to vinnv226's post “By definition, "energy de...”
  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.
  Comment on vinnv226's post “By definition, "energy de...”
  (31 votes)
WAJIH RIZVI
Posted 7 years ago. Direct link to WAJIH RIZVI's post “I don't understand what n...”
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?
Button navigates to signup pageButton navigates to signup page
(10 votes)
Answer
- Ishaan Sahasrabudhe
  Posted 4 years ago. Direct link to Ishaan Sahasrabudhe's post “Net-work just means the _...”
  Net-work just means the total work done by the system. And the value of net work = change in kinetic energy.
  Comment on Ishaan Sahasrabudhe's post “Net-work just means the _...”
  (21 votes)
😊
Posted 7 years ago. Direct link to 😊's post “prove that K.E = 1/2 mv2?”
prove that K.E = 1/2 mv2?
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(3 votes)
Answer
- 🚀The knowledge Hunter🔭
  Posted 7 years ago. Direct link to 🚀The knowledge Hunter🔭's post “You are asking a question...”
  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
  Comment on 🚀The knowledge Hunter🔭's post “You are asking a question...”
  (18 votes)
Kyle Delaney
Posted 7 years ago. Direct link to Kyle Delaney's post “Why is kinetic energy rel...”
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?
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- blancaroca60
  Posted 6 years ago. Direct link to blancaroca60's post “The simplest derivation i...”
  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.
  Button navigates to signup page
  (5 votes)
GiGI25
Posted 3 years ago. Direct link to GiGI25's post “if there is the another w...”
if there is the another way to say kinetic energy
Button navigates to signup pageComment on GiGI25's post “if there is the another w...”
(7 votes)
Answer
- 〰〰Cadence〰〰
  Posted 2 months ago. Direct link to 〰〰Cadence〰〰's post “There are multiple, but k...”
  There are multiple, but kinetic energy is the most appropriate term :)
  Button navigates to signup page
  (1 vote)
muhammad
Posted 5 years ago. Direct link to muhammad's post “if the propellant is cont...”
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?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Andrew M
  Posted 5 years ago. Direct link to Andrew M's post “The propellant is used up...”
  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)
  Comment on Andrew M's post “The propellant is used up...”
  (6 votes)
jacobkwak9293
Posted 3 years ago. Direct link to jacobkwak9293's post “For exercise 2, I want to...”
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?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Ahmad Abu Zainab
  Posted 3 years ago. Direct link to Ahmad Abu Zainab's post “Yes, you should convert M...”
  Yes, you should convert MJ to J. The reason that your answer must me multiplied by 1000 since sqrt(10^6)=10^3=1000
  Button navigates to signup page
  (4 votes)
Thermos
Posted 6 years ago. Direct link to Thermos's post “Can you say that kinetic ...”
Can you say that kinetic energy is the source of sound energy? Can you say that kinetic energy and sound energy exist together?
Button navigates to signup pageComment on Thermos's post “Can you say that kinetic ...”
(4 votes)
Answer
- Darth Skytron
  Posted 5 years ago. Direct link to Darth Skytron's post “Sound energy is a type of...”
  Sound energy is a type of kinetic energy.
  Button navigates to signup page
  (3 votes)
Ada Li
Posted 4 years ago. Direct link to Ada Li's post “Why does motion need ener...”
Why does motion need energy?
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(4 votes)
Answer
- gagste001
  Posted 4 months ago. Direct link to gagste001's post “there would be no motion ...”
  there would be no motion with out energy
  Button navigates to signup page
  (1 vote)
Jay Malave
Posted 6 years ago. Direct link to Jay Malave's post “If you see in first parag...”
If you see in first paragraph,how will object move with a constant speed if I apply
energy only once? It will stop after a certain interval of time
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Andrew M
  Posted 6 years ago. Direct link to Andrew M's post “If it stops, that means s...”
  If it stops, that means someone or something else was "applying energy". In physics we call that "doing work". Friction does work to stop the object after you did work to get it moving. If you did it in outer space the object wouldn't stop
  Button navigates to signup page
  (6 votes)

Physics library

Course: Physics library > Unit 5

What is kinetic energy?

What is kinetic energy?

How can we calculate kinetic energy?

What is interesting about kinetic energy?

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