# How would you calculate the work done

1. Jan 11, 2006

### stunner5000pt

if the force depended on the DIRECTION of the VELOCITY
that is $$\vec{F} = - F \hat{v}$$.
so suppose it was dragged along a path - say from (0,0) to (1,1) i na straight line what is the work done?
It going in a striaght line so we could parametrize the curve into r(t) = (t,t)
and r'(t) = (1,1) right
and thus v hat = (1,1) as well then?
now the problem is the work depends on the direction of the velocity

so then the work done is $$W = -F \int_{0}^{1} (1,1) (1,1) dt$$ is that right?

ok so lets say it was dragged along the curve y = x^2
then r(t) = (t,t^2)
and r'(t) = (1,2t)
and $$\hat{v} = \frac{(1,2t)}{\sqrt{1+4t^2}}$$
ok but what the dirrefetial quantity fo the function here? Is it (t,t^2) dt?
so then is
$$W = -F \int_{0}^{1} \frac{(1,2t))}{\sqrt{1+4t^2}} (t,t^2) dt$$
is that right?

2. Jan 11, 2006

### StatusX

You should use:

$$W=\int \vec F \cdot d\vec x = \int \vec F \cdot \vec v \mbox{ } dt$$

If $\vec F = -|F| \hat v$, then:

$$W= \int -|F|\hat v \cdot \vec v \mbox{ } dt = -|F|\int |v|\hat v \cdot \hat v \mbox{ } dt$$

$$= -|F| \int |v|dt = -|F| \int ds$$

So you don't have to do most of that work you did.

Last edited: Jan 11, 2006
3. Jan 11, 2006

### stunner5000pt

oh ok that seems much simpler
but was i did correct by teh way?

doing it the simpler way though...
for the straight line from (0,0) to (1,1)
the integrand would be
$$\int_{0}^{1} \int_{0}^{1} (1,1) \bullet (1,1) dx dy$$ ?
and for hte line from (0,0) to (1,1) along y = x^2
$$\int_{0}^{1} \ int_{y=0}^{y=x^2} (1,1) \bullet (1,1) dy dx$$

are thos correct?

4. Jan 11, 2006

### StatusX

No, there aren't any double integrals in this. 'ds' means the differential arclength. Note that I substituted it in for |v|dt. The integral of ds over a segment of a curve is the length of that segment of the curve. For a curve y(x) in 2D, ds2=dx2+dy2=dx2(1+(dy/dx)2). For example, for the straight line from (0,0) to (1,1), $\int ds =\sqrt{2}$.

Last edited: Jan 11, 2006
5. Jan 11, 2006

### stunner5000pt

ooooo i see i understnad now
there was another part of this problem that involved proving what you proved in your first post. I totally disregaraded the meaning of dS. SO the integral of dS represents the arc length.

so for the y=x^2 function the arc legnth would be for r(t) = (t,t^2) and r'(t) = (1,2t)
$$\int_{0}^{1} \sqrt{1 + 4t^2} dt$$ yes?

6. Jan 12, 2006

### stunner5000pt

can i get a confirmation on my answer? Is it right? I m not sure on how to integrate it though. I tried using mathematica and got an answer that involved sinh which i haven ot come across as of yet.