I am wondering if anyone can help to let me know if I am on the right track or making mistakes. I am really not so confident in my work here so I am really looking for someone to look over it. It is the following problem: Yet again this is another problem I have tried significantly on my own and by trying to show my work on here, yet have received no full answers or even advice to incorporate the friction
A point is attracted to the origin by a force that is proportional to the cube of the distance from the origin, how much work is done moving the point from O to $(2,4)$ along the path $y=x^2$ assuming we have some coefficient of friction $\mu$
First question: Will I need a positive or negative sign infront of the force? I assume since it is attracted to the origin and we are moving away from the origin wouldn't we need a positive? But I am not sure.
I think my Force would be $$F=-C(x^2+y^2)^{3/2} \frac{OP}{||OP||}$$=$$-C(x^2+y^2)^{3/2}\frac{xi+yj}{(x^2+y^2)^{1/2}}$$ =$-C(x^2+y^2)(ti+tj)$ Does that seem okay? Should I have a minus sign in front of the C?
So for the force before considering friction,
parametrized with $r(t)=(t,t^2)$, $r'(t)=(1,2t)$ with $0 \le t \le 2$
$F(r(t))=-C(t^2+t^4)(ti)+(t^2+t^4)(tj)$
$F(r(t)) \bullet r'(t)= -C(2t^7+3t^5+t^3)$
integrated from $0$ to $2$, gives $$-100C$$ But I am having the more trouble with the friction part.
Moving on to the friction component, I know that work against friction is given by $-\mu | \int_{c} F_{N}dS|$
My unit tangent T is $$(\frac{1}{\sqrt{1+4t^2}},\frac{2t}{\sqrt{1+4t^2}})$$
and so would by unit Normal just be
$$ N= ( \frac{-2t}{(4t^2+1)^{3/2}(\sqrt{1/(4t^2+1)}},\frac{1}{(4t^2+1)^{3/2}(\sqrt{1/(4t^2+1)}} $$
and then If I take the dot product with F, and integrate that result, that will give the work by friction? And then I use Net Work= Work from central force+ W from friction? But since the form of the normal is so messy I am not even sure how to do the dot product in the correct way. and how to write F in such a way to do the dot product. I know it is done by component, but in my F I have a xi+yi, etc
I am still looking for help on incorporating the frictional component on this problem.
Thank you
Work done by the attractive force
The work done by the attractive force $\newcommand{F}{\mathbf F}\F$ along the path $C$ is $$ \newcommand{T}{\mathbf T} W_\F = \int_C \F \cdot \T \;ds $$ where $\T$ is the unit tangent vector and $s$ is distance along the curve. If we write $\newcommand{r}{\mathbf r}\r$ for the radial (position) vector of the particle, $r = \lvert\r\rvert$ for the distance from the center, and $\hat\r = \frac1r \r$ for the unit radial vector, then the force $\F$, which is a central force, can be written in the form $\F = f(r) \,\hat\r$. Using the fact that $\frac{dr}{ds} = \hat\r \cdot \T$, $$ W_\F = \int_C f(r)\, \hat\r \cdot \T \;ds = \int_C f(r)\, \frac{dr}{ds} \;ds = \int_C f(r)\, dr = \int_{r_0}^{r_1} f(r)\, dr \tag1 $$ where $r_0$ and $r_1$ are the distances from the center to the starting and ending points of $C$.
In other words, $\F$ is a conservative force whose potential is a function of distance from the center. It's useful to be able to recognize that a force is conservative, because it frees us from having to trace the work done along a particular path. In this problem, we simply set $f(r) = -kr^3$, $r_0 = 0$, and $r_1 = \lVert (2,4) \rVert = 2\sqrt5$, and integrate \begin{align} W_\F &= \int_0^{2\sqrt5} -kr^3\, dr \tag2\\ &= -k \left[ \frac14 r^4 \right]_0^{2\sqrt5} = -k\left(\frac14\left( 2\sqrt5 \right)^4\right)\\ & = -100k. \end{align} (I used $k$ rather than $C$ as the constant of proportionality because we already used $C$ to name a curve.)
Of course this is the same result you found (except for the choice of the name of the constant), and I knew that when I wrote this up. So why bother with all this? Several reasons:
It's one more confirmation that you had that part of the solution already.
Equation $(1)$ helps solve many problems, not just this one. It's a general property of force fields that (like this one) have spherical symmetry. If you are dealing with such a field you can pretty much just write down something like Equation $(2)$ and evaluate it.
It demonstrates a way to turn a path integral into an ordinary definite integral via a change of variables without having to decide in advance how you want to parameterize your curve.
Work done by the frictional force
The frictional force is not conservative, so we still have to integrate it over the specific path $C$.
Note that I wrote $(2,4)$ for the vector from the point $(0,0)$ to the point $(2,4)$. An alternative notation is $\newcommand{ihat}{\hat \imath}\newcommand{jhat}{\hat \jmath} 2\ihat + 4\jhat$, where $\ihat = (1,0)$ and $\jhat = (0,1)$ are an orthonormal basis of this two-dimensional vector space. In other words, when you wrote (with slightly different symbols) $$ \F = -k(x^2 + y^2)(x\ihat + y\jhat), $$ it's the same vector as $\left(-k(x^2+y^2)x,-k(x^2+y^2)y\right)$, which can also be written as the product of the scalar $-k(x^2+y^2)$ with the vector $(x,y)$: $$ \F = -k(x^2 + y^2)(x, y). $$ On the path $C$ defined by $y = x^2$, this is $$ \F = -k(x^2 + x^4)(x, x^2). \tag3 $$
Conversely, your unit tangent vector, $$\T = \left( \frac{1}{\sqrt{1+4x^2}}, \frac{2x}{\sqrt{1+4x^2}} \right),$$ could instead be written $$\T = \frac{1}{\sqrt{1+4x^2}} \ihat + \frac{2x}{\sqrt{1+4x^2}} \jhat.$$
Admittedly, it's a little confusing to mix and match these two notations when writing an inner ("dot") product of two vectors, so just convert both vectors to the same notation and evaluate the inner product in that notation: either add up the componentwise products (if you use component notation), or use the facts that $\ihat\cdot\ihat = \jhat\cdot\jhat = 1$ and $\ihat\cdot\jhat = 0$.
For your normal vector, you simply need to rotate the tangent vector through a right angle. In general, a right-angled rotation of the vector $(u,v)$ will give you either the vector $(-v,u)$ or the vector $(v,-u)$; in this problem, the unit normal vector you want is the one that points to the "downward" side of the path (closer to the origin), so it is $$ \newcommand{N}{\mathbf N} \N = \left( -\frac{2x}{\sqrt{1+4x^2}}, \frac{1}{\sqrt{1+4x^2}} \right) $$ (although you could use either vector since we're going to take an absolute value later), but I think it's even easier to use the fact that $\frac{ds}{dx} = \sqrt{1+4x^2}$ to write the vector $\N$ as a scalar multiple of the vector $(-2x,1)$: $$ \N = \left(\frac{dx}{ds}\right) (-2x, 1). \tag4 $$
If you use Equations $(3)$ and $(4)$ to make substitutions for $\F$ and $\N$ in the formula for the work done on the particle by friction, $$ W_{\mathrm {friction}} = -\mu \int_C \lvert \F \cdot \N \rvert \;ds, $$ the scalar factor $\frac{dx}{ds}$ allows a convenient change of integration variable (as $\frac{dr}{ds}$ did before), and you're left with a scalar multiplied by the inner product $(x,x^2)\cdot(-2x,1)$ in the integrand: \begin{align} W_{\mathrm {friction}} & = -\mu \int_C \left\lvert -k(x^2 + x^4)(x, x^2) \cdot \left(\frac{dx}{ds}\right) (-2x, 1) \right\rvert \;ds \\ & = -\mu \int_C k(x^2 + x^4) \;\left\lvert(x, x^2) \cdot (-2x, 1)\right\rvert\; \frac{dx}{ds}\,ds \\ & = -\mu \int_0^2 k(x^2 + x^4) \;\left\lvert(x, x^2) \cdot (-2x, 1)\right\rvert\; dx \\ \end{align} The rest is just a matter of the usual simplifications and evaluations; here's the final result in Wolfram Alpha to check your work against.
By the way, using the scalar factor $\frac{dx}{ds}$ in $\N$ was not just an inspired guess or some weird stroke of genius. It was simply motivated by really not wanting to use $s$ as the variable of integration. So I looked around at what the derivatives of more convenient variables such as $r$, $x$, or $y$ would be and whether I could turn some part of the integrand (especially an othewise inconvenient part like the factor of $1/\sqrt{1+4x^2}$) into such a handy differential so that I could make a change to a more convenient variable of integration.