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Alabaster
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Homework Statement
The kinetic energy K of an object of mass m and velocity v is given by K=1/2mv2. Suppose an object of mass m, moving in a straight line, is acted upon by force F=F(s) that depends on position s. According to Newton's Second Law F(s)=ma. A is acceleration of the object.
Show that the work done from moving a position s0 to a position s1 is equal to the change in the object's kinetic energy; that is, show that
Work=[itex]\int^{s1}_{s0}[/itex]F(s)ds=[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{1}[/itex]-[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{0}[/itex]
where v0=v(s0) and v1=v(s1) are the velocities of the objects at positions s0 and s1.
Homework Equations
I think all of the relevant equations are listed above.
The Attempt at a Solution
Work=[itex]\int^{s1}_{s0}[/itex]F(s)ds=[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{1}[/itex]-[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{0}[/itex]
Looking at the right side of the equation suggests that I take the integral and it splits into two, since it's a definite integral. So I should start with this Work=[itex]\int^{s1}_{s0}[/itex]F(s)ds and mess with it until it's the same as the right side of the equation.
This is where I get stuck. I asked a guy for help and he showed me
ma=m[itex]\frac{dv}{dt}[/itex]
and then said "s" is a displacement vector s=vdt
He showed me the next step, which was =[itex]\int^{t1}_{t0}[/itex]Fvdt
I'm very confused how he got there, but he showed me the whole problem. This is it in its entirety.
Work=[itex]\int^{s1}_{s0}[/itex]F(s)ds
=[itex]\int^{t1}_{t0}[/itex]Fvdt
=[itex]\int^{t1}_{t0}[/itex]m[itex]\frac{dv}{dt}vdt[/itex]
=m[itex]\int^{t1}_{t0}[/itex]vdv
=m(v2/2) evaluated from t0 to t1
=[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{1}[/itex]-[itex]\frac{1}{2}[/itex]mv[itex]^{2}_{0}[/itex]
If someone could explain what happened step-by-step, that would be amazing. I can follow some of the steps, but all the variable swapping is making this impossible for me to understand.
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