Differential equations and physics proof problem

[Alabaster]

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=\int^{s1}_{s0}F(s)ds=\frac{1}{2}mv^{2}_{1}-\frac{1}{2}mv^{2}_{0}

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=\int^{s1}_{s0}F(s)ds=\frac{1}{2}mv^{2}_{1}-\frac{1}{2}mv^{2}_{0}

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=\int^{s1}_{s0}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\frac{dv}{dt}
and then said "s" is a displacement vector s=vdt

He showed me the next step, which was =\int^{t1}_{t0}Fvdt

I'm very confused how he got there, but he showed me the whole problem. This is it in its entirety.

Work=\int^{s1}_{s0}F(s)ds

=\int^{t1}_{t0}Fvdt

=\int^{t1}_{t0}m\frac{dv}{dt}vdt

=m\int^{t1}_{t0}vdv

=m(v²/2) evaluated from t₀ to t₁

=\frac{1}{2}mv^{2}_{1}-\frac{1}{2}mv^{2}_{0}

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.

 

[tjackson3]

You've got your spatial integral bounds in the wrong order - you should go from s0 to s1.

So the first step is to note that F(s) = ma, as you have. Velocity is the derivative of position: v = ds/dt; a = dv/dt, so F(s) = m dv/dt. The first trick is that we need to convert this spatial integral into a time integral. Since v = ds/dt, v dt = ds, which is what allows you to make the change of variables that seems to have confused you.

 

[Alabaster]

You were correct, my bounds were wrong and I fixed them. And thank you for responding. This little bit you told me helps so much. It's late here, but I'll definitely take another look at this tomorrow.

 

[Alabaster]

I've pretty much gotten this figured out at this point. I still have a couple questions at the end, though.=m\int^{t1}_{t0}vdv

After the dt's cancel each other, the problem has a interval with the upper and lower bounds of t, but there's a dv now that the integral works with. Does that effect the integration in any way? Should I now be evaluating with the bounds of v instead of t?

=m(v²/2) evaluated from t₀ to t₁

At this point, how do the t's fit into the problem? Normally you'd replace a variable with the upper bound and subtract it by the replaced lower bound of the same equation, but what do I do with the upper and lower bound variables on my integral symbol?