# I Work-Energy theorem if the particle´s mass depends on time

1. May 21, 2017

### vdvdlk

This weekend I was trying to calculate the work-energy theorem, considering a body that can be treated like a particle, and has its mass varying in time. I searched through a lot of sites if such thing existed, and didn´t find anything. Then I found a thread (https://www.physicsforums.com/threads/work-energy-theorem-for-variable-mass-systems.720649/) from 2013, asking that same thing, but appeared to not be answered. This motivated me to put my results in here. As I´m not familiarized with writing the math in the computer, I put the image of how I did it below. Summarizing, I found this equation:

dW = dK + (1/2)v2dm

Where W stands for the work done by the net force F, K stands for kinectic energy, m stands for mass in function of time, and v stands for the norm of the velocity.

http://imgur.com/a/MUhaT

Is this result correct? If not, where did I go wrong?

2. May 22, 2017

### DrStupid

3. May 22, 2017

### AlphaLearner

No, you and @vdvdlk are wrong. Because, varying mass will even lead to varying velocities! When mass increase, velocity decrease. When mass decrease, velocity increase just to put up that total KE value constant.
But you can still study its change in mass and velocity, but you have to make up the formula.

Recipe: Knowledge upon momentum in variable mass system.

Skills/Techniques to make it: Mathematics, solving equations using algebra and some calculus.

Material Source: Click Here. Read only the first part, It is the concept. Second part is all about its applications in rockets (Up to you even if you read or not)

Both momentum with varying mass and kinetic energy with varying mass have a same concept: The total momentum/kinetic energy always remain constant by shifting velocity as mass shift.

Then why momentum first? Because, momentum with varying mass exists readymade explained in materials and online sources like I have given one and kinetic energy does not (As you say)

When you get familiar with momentum with varying mass, you find it easy to apply even in kinetic energy.

Just give it a try and put your final equation from your work and conclude in this thread. Even I put up my result tomorrow upon solving it on my own.
Good Luck!

4. May 22, 2017

### DrStupid

That depends on the circumstances. The velocity of a rocket doesn't need to in crease if it's mass decreases and the velocity of a ballistic pendulum doesn't need to decreases when it's mass increases and the KE of an open system doesn't need to be constant.

$K = {\textstyle{1 \over 2}}m \cdot v^2$
$dK = m \cdot v \cdot dv + {\textstyle{1 \over 2}}v^2 \cdot dm$
$dW = F \cdot ds = m \cdot \dot v \cdot ds + v \cdot \dot m \cdot ds = m \cdot v \cdot dv + v^2 \cdot dm = dK + {\textstyle{1 \over 2}}v^2 \cdot dm$

5. May 23, 2017

### AlphaLearner

Since you have mentioned 'Open System', your statement can be agreed to an extent. But in closed rocket - earth system, when mass of rocket decrease as it eject out fuel as it goes higher, velocity must and should increase. But due various external forces like gravitational force or air resistance acting upon it, You may not find it true. But in a space where the particle is not affected by any external forces, the the particle definetly shift its velocity with respect to change in mass in order to keep the value of the Kinetic Energy constant. But for now, for @vdvlk discussing regarding KE in closed system is necesary to understand than in an Open system which makes it more complicated for him.

I'm still not done with the solving part since I didn't find time for it... Sorry for the delay. If a mentor comes into this thread, He can surely help you by verifying your equation or even mine (when kept later).
Your equation clearly show change in velocity and you dont say it happens so...

6. May 23, 2017

### DrStupid

As it was clear from the beginning that we are talking about open systems, this means that your original statement is wrong.

No, it doesn't. The speed of the accelerating rocket can also decrease or remain constant - even without external forces.

No, it doesn't. There is no reason for the kinetic energy of a particle with changing mass to remain constant.

7. May 23, 2017

### vdvdlk

What do you guys mean by a closed or open system?

8. May 23, 2017

### DrStupid

Closed systems can interact with the environmend but don't exchange mass. Open systems also exchange mass.

9. May 24, 2017

### AlphaLearner

Well... once again, 'Open System' has all to do with what you have said. But just tell me one thing, will you begin topics like newton's laws of moton, work and energy, system of particles, momentum, gravitation, thermodynamics and to say complete mechanics with an 'Open System' or 'Closed system'? Which one sounds easy to study first. Dont talk about 'Open Systems' unless this topic gets clear with 'Closed systems'.

This topic is not intented to talk about 'Open' or 'Closed' systems. If, any of us can frame the equation for it as correct as possible, we will give it off to @vdvlk.

10. May 24, 2017

### DrStupid

Closed systems are off-topic.

11. May 24, 2017

### AlphaLearner

What do you mean? Dont you study simple systems rather go for more complex systems where energy and forces exist in vast forms which beginners (Including me to say) dont know over more than 90% what they are and where they come from too? It's like literally scaring off beginners away from physics. Consider, Kinetic Energy of a particle moving in free space is defined as KE = 1/2mv2 in a closed, conserved system. When expose that same partilcle to an 'Open system', you should start considering effects, Air resistance, Gravity (Leading it to projectile motion than straight path) etc. etc. and when it comes to apply it all in a simple equation like 1/2mv2, It is a completely huge task to get even such a simple result. Try that yourself?

12. May 24, 2017

### DrStupid

The mass of closed systems is invariant in classical physics. This thread is about systems with time-dependend mass.

The OP and I already did it and the calculation appears to be quite simple (see above). When do you plan to contribute something useful to this thread?

Last edited: May 24, 2017