What is the Relationship Between Force, Power, and Work in a Particle System?

In summary, the conversation discusses the calculation of work and power in a scenario where a force is acting on a particle and its movement is being integrated. The participants consider different approaches, including using the equation W = \int Fdx and calculating instantaneous power using the equation P = Fv. They also mention the importance of considering the boundaries and adapting for the velocity when calculating power.
  • #1
sAXIn
12
0
I have a question where I have a force F(t) acting on particle ,
I intgrate 2 times and find X(t) , i used innitial conditions given x=0 , t=0 , v=0 .

1) now they ask me to write the work done by the power between t=0 , t=<epsilon> I know that dw=F dot DX but here the f and x are both functions of t so ... ?

2) after that i need to write the power of the force on time t>0 i think interval t=0 , t=t ... ?

thank's in advance
 
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  • #2
1)
You have dx/dt = ... -> dx = ...dt
[tex]W = \int_{x_0}^{x}Fdx = \int_{t_0}^{t}F(t)...dt[/tex]

You know F in terms of t to get the solution.

2)
P = Fv = F(t)v(t), which you already have.
 
  • #3
yes but you don't put in consideration that x is function of t it\s integral of F(t)d(x(t)) or something

2) I know that p=FV when the force is constant here it isn't ?!
 
  • #4
sAXIn,

There are two options actually.

Suppose (i take x and F both along the x-axis, but you know that the integrandum of the work is a SCALAR product of two vectors)
F = 2
x = t²

You just need to substitute everything towards t...
Then [tex]W = \int Fdx = \int 2d(t^2) = \int 4tdt[/tex]

Make sure that you adapt the boundaries.

or, since you know the velocity , you can calculate the kinetic energy and you know that work is equal to the difference in kinetic energy between the end and beginning of the motion : [tex]W = E_{k}^{FINAL} - E_k^{BEGINNING}[/tex]

marlon
 
  • #5
1) Of course it is used:

[tex]W = \int_{t_0}^{t}F(t)v(t)dt[/tex]

where dx = v(t)dt. So what's the problem.

2)
You need instantaneous power, which is the product of instantaneous force at time t and instantaneous velocity at time t. If you had used

[tex]P = \frac{\Delta W}{\Delta t}[/tex]

then it would have been average power that doesn't apply in this problem.

If you're skeptic:

[tex]P = \frac{d}{dt}[W(t)] = \frac{d}{dt}(\int_{}^{}F(t)vdt) = Fv[/tex]


where you well know that dx = vdt.
 
  • #6
I got it's the same : I can integrate f(t)v(t)and then def. by dt so its = power or just skip it and f(t)v(t) is my power ! thanks a lot ...
 

Related to What is the Relationship Between Force, Power, and Work in a Particle System?

1. What is the difference between force and power?

Force is a physical quantity that causes an object to accelerate or change its motion, while power is the rate at which work is done or energy is transferred.

2. How do you calculate work?

Work is calculated by multiplying the force exerted on an object by the distance the object moves in the direction of the force. The formula for work is W = F x d, where W is work, F is force, and d is distance.

3. What are the units of force, power, and work?

The SI unit for force is Newton (N), for power is Watt (W), and for work is Joule (J).

4. Can work be negative?

Yes, work can be negative. This occurs when the force and the displacement are in opposite directions, resulting in a negative value for work.

5. How is power related to work and time?

Power is equal to the amount of work done divided by the time it takes to do the work. The formula for power is P = W/t, where P is power, W is work, and t is time.

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