Proving the Work-Energy Theorem A lot of calculus, anyone?

In summary, the conversation discusses a particle moving on a curved path and being acted upon by a net force. The goal is to prove the work energy theorem for this general case. This involves using the equations F=ma and W=Fd, and integrating to show that the total work done is equal to the change in kinetic energy. The conversation also mentions the need for parametrization and understanding of line integrals.
  • #1
33
0

Homework Statement



A particle moves on a curved path from (x1, y1,z1) to (x2, y2,z2). At the start, the particle has a velocity of v = v1xi+v1yj+v1zk. This curved path can be divided into segments infinitesimally, which are, dl = dxi +dyj +dzk. It is acted on by a net force F = Fxj + Fyi + Fzk. The force components Fx, Fy, and Fz, are in general functions of position. Prove the work energy theorem for this general case. That is prove that

Wtot=K2-K1

where

Wtot=[itex]\int[/itex]F dl = [itex]\int[/itex] Fxdx + Fydy + Fzdz

where the limits of integration are from (x1, y1,z1) to (x2, y2,z2) for each.

Homework Equations



W = Fd
∫F=W

ax= [itex]\stackrel{dvx}{dt}[/itex] = [itex]\stackrel{dvx}{dx}[/itex] *[itex]\stackrel{dx}{dt}[/itex] = vx[itex]\stackrel{dvx}{dx}[/itex]

The Attempt at a Solution




I don't really have any idea where to start but what I did was I took the velocities and made them into accelerations then changed the Fx, Fy, and Fz, into max, may, and maz values, then I'm confused now because I have an integral which looks like this..

[itex]\int[/itex] m*vx*dvx + m*vy*dvy/dx dy + m*vz*dvz/dx dz

which makes no sense because the dx only canceled out for the X direction, and I need to prove that Wtot = 1/2 mvf^2 - 1/2mvi^2


If anyone could help it'd be appreciated because I've got really no idea what I am doing
 
Physics news on Phys.org
  • #2
Bolded means vectors. Note that x = (x1,x2,...xn). The period denotes dot product.

Let W = ∫F.dx
F = m(d2x/dt2)

W = m∫(d2x/dt2).dx
W = m∫(dv/dt).dx

Note that this step is somewhat hand-wavy but it's true. I think it would be a good exercise for you to prove this. To do that, you first need to learn about how line integrals work, and the process of parametrization. Hint: dx=(dx/dt)dt

W = m∫dv.(dx/dt)
W = m∫(dx/dt).dv
W = m∫v.dv
W = m(1/2 v.v) = 1/2 m|v|2

Is your name Ping?
 
Last edited:

1. What is the Work-Energy Theorem?

The Work-Energy Theorem is a fundamental principle in physics that states the net work done on an object is equal to the change in its kinetic energy.

2. How is the Work-Energy Theorem mathematically represented?

The Work-Energy Theorem can be mathematically represented as W = ΔKE, where W is the net work done on an object and ΔKE is the change in kinetic energy of the object.

3. What is the role of calculus in proving the Work-Energy Theorem?

Calculus is essential in proving the Work-Energy Theorem because it allows us to calculate the work done on an object by integrating the force over a distance, and to calculate the change in kinetic energy by integrating the object's acceleration over time.

4. Can the Work-Energy Theorem be applied to all types of motion?

Yes, the Work-Energy Theorem can be applied to all types of motion, including linear, rotational, and oscillatory motion. It is a universal principle that applies to all objects in motion.

5. How is the Work-Energy Theorem used in real-world applications?

The Work-Energy Theorem is used in many real-world applications, such as calculating the energy output of a machine, determining the efficiency of a system, and understanding the motion of objects in various scenarios, such as car crashes or roller coaster rides.

Suggested for: Proving the Work-Energy Theorem A lot of calculus, anyone?

Replies
11
Views
881
Replies
9
Views
1K
Replies
23
Views
979
Replies
9
Views
919
Replies
3
Views
936
Replies
14
Views
2K
Back
Top