Prooving Kinetic Energy is integration of Work. I have a little problem?
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So here goes. I was trying to proove that KE is the int. of work to my teacher but he told me what I was doing isnt right, so here is what i did. int.Work=KE Work=int(F.dx) Work=int(ma.dx) work =m int((dv/dt)*dx) now here is the problem. What i did was I switched the dv/dt and dx. I switched the dv/dt into dx/dt*dv. Which would make it V(dx/dt)dv, and if you integrate Vdv i would get the Ke formula. So what my teacher said is that I cannot do what i did as in i cannot switch the dv/dt and dx places. Is there any other way of doing this. He told me this actually takes a page of calculus to explain. Pls help
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Answer:
First law of thermodynamics says : The sum of work and heat is equal to the total change in energy of a mass. In some special cases, work equals the change in kinetic energy and not what you're trying to prove, and one more thing, you can't switch the dv/dt into (dx/dt)*dx that way.
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Other answers
First law of thermodynamics says : The sum of work and heat is equal to the total change in energy of a mass. In some special cases, work equals the change in kinetic energy and not what you're trying to prove, and one more thing, you can't switch the dv/dt into (dx/dt)*dx that way.
Cool!!!
No, you cannot switch dv/dt with x, because dv/dt is the change in velocity over the change in time, which is acceleration, not displacement. Velocity is change in distance over time, or dx/dt, which is present in your equation. So dx/dt yields v, leaving your equation to be... work =m int(dvdx/dt) work = m int(v dv) Now integrate v with respect to v... work = m(v^2/2) Or... (1/2)mv^2, which is the equation for KE. Although with limits of integration, your full equation is... KE = (1/2)m(v final)^2 - (1/2)m(v initial)^2 Hope I helped.
Mind Flayer
My dear you are wrong. Read your physics book.
Fazaldin A
No, you cannot switch dv/dt with x, because dv/dt is the change in velocity over the change in time, which is acceleration, not displacement. Velocity is change in distance over time, or dx/dt, which is present in your equation. So dx/dt yields v, leaving your equation to be... work =m int(dvdx/dt) work = m int(v dv) Now integrate v with respect to v... work = m(v^2/2) Or... (1/2)mv^2, which is the equation for KE. Although with limits of integration, your full equation is... KE = (1/2)m(v final)^2 - (1/2)m(v initial)^2 Hope I helped.
Mind Flayer
My dear you are wrong. Read your physics book.
Fazaldin A
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