1D (net) work done by (net) force on a variable mass system.

In summary, the conversation discusses the relation between mechanical work and kinetic energy in a system of constant mass. The equation W = ΔK = ½mΔv2 is used as an example, but the question of how this applies to a system with varying mass is then raised. The speaker attempts to derive an expression for this scenario and considers a box with varying mass on a frictionless table being pushed by a force that also varies through time. The equation for Newton's second law is used to calculate the work done by the force, and the speaker questions the validity of their reasoning.
  • #1
da_nang
137
5
So I was sitting on the train last weekend, reading through my physics book on mechanical work and its relation to kinetic energy. One example would be that a box on a frictionless table being pushed and they would conclude that W = ΔK = ½mΔv2.

Looking at this equation got me thinking, obviously this applies to a system of constant mass, but what if the mass varied? I tried looking it up on the Internet, but either I'm wording it wrong or there isn't much info out there freely available. So I figured I should try to derive an expression on my own.

Now then, consider a box of some mass on a frictionless table in vacuum. Then let's say a piston pushes the box from rest with a force that varies through time i.e. [tex]\overrightarrow{F} = \overrightarrow{F}(t)[/tex]

Now since the mass isn't constant, we'll use a more general equation for Newton's second law of motion: [tex]\overrightarrow{F}(t) = \frac{d(m\overrightarrow{v})}{dt} = \frac{dm}{dt}\overrightarrow{v} + \frac{d\overrightarrow{v}}{dt}m[/tex] where mass and velocity are functions of time.

Then, consider that the work done by a force is [tex]W = \int \overrightarrow{F}\cdot d\overrightarrow{s}[/tex] or in this case: [tex]W_{F}(t) = \int \overrightarrow{F}(t)\cdot d\overrightarrow{s}[/tex]

In this 1D problem, the force acts in the same direction as the displacement, so the angle between the vectors are zero and so we can write the equation thusly:

[tex]W_{F}(t) = \int F(t)ds[/tex] Expanding F(t):

[tex]W_{F}(t) = \int v\frac{dm}{dt}ds + \int m\frac{dv}{dt}ds[/tex] Using the chain rule and cancelling ds:

[tex]W_{F}(t) = \int v\frac{dm}{ds}\frac{ds}{dt}ds + \int m\frac{dv}{ds}\frac{ds}{dt}ds = \int v^{2}dm + \int mvdv = mv^{2} + \frac{1}{2}mv^{2} + C = \frac{3}{2}mv^{2} + C[/tex] Since the box was at rest in the beginning, the constant C = 0.

[tex]W_{F}(t) = \frac{3}{2}mv^{2} \left( = \frac{3}{2}m(t)v(t)^{2}\right)[/tex]

So my question is, is this equation valid classically? Have I made any mistakes in my reasoning?

Thanks in advance.
 
Physics news on Phys.org
  • #2
The integrals are not valid-- you did the integral over dm as if v was constant, and the integral over vdv as if m was constant. Where it gets interesting is if you imagine an object moving with constant velocity that starts out with negligible mass an gradually accumulates to mass m, maintained at v by a force. You then can take just your first integral to show that the work done by the force is mv2, even though the final kinetic energy is 1/2 that. The remaining 1/2 is thermalized internally when you stick on mass, i.e., it goes into heat.
 

1. What is 1D (net) work done by (net) force on a variable mass system?

The 1D (net) work done by (net) force on a variable mass system refers to the amount of energy transferred to or from a system in one direction by the combined effect of all forces acting on the system. It takes into account both the magnitude and direction of the forces, as well as the change in mass of the system.

2. Why is 1D work done by force important in a variable mass system?

1D work done by force is important in a variable mass system because it helps to determine the overall energy balance of the system. It can also be used to calculate the change in kinetic energy of the system, which is essential in understanding the motion and behavior of the system.

3. How is the 1D work done by force calculated in a variable mass system?

The 1D work done by force can be calculated by multiplying the magnitude of the force by the displacement of the system in the direction of the force. It can also be calculated as the change in kinetic energy of the system, which can be determined by the mass and velocity of the system at two different points in time.

4. What factors can affect the 1D work done by force in a variable mass system?

The 1D work done by force can be affected by several factors, including the magnitude and direction of the force, the displacement of the system, and the change in mass of the system. Other factors such as friction, air resistance, and external forces can also affect the work done by force on the system.

5. How is the concept of 1D work done by force applied in real-world situations?

The concept of 1D work done by force is applied in various real-world situations, such as in the motion of vehicles, the launching of rockets, and the operation of machines. It is also used in fields such as engineering, physics, and biomechanics to analyze and understand the behavior of systems under the influence of forces.

Similar threads

  • Mechanics
Replies
30
Views
761
Replies
17
Views
887
Replies
0
Views
1K
  • Mechanics
Replies
3
Views
912
  • Mechanics
Replies
6
Views
807
Replies
14
Views
1K
Replies
35
Views
3K
  • Mechanics
Replies
20
Views
864
Replies
43
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
1K
Back
Top