# work done using line integral

Question :

How much work is done in pulling an object constrained to move along the portion of the curve y = $x^2$; z = $x^3$ from (0; 0; 0) to (1; 1; 1) (positions in meters), if the rope pulling it is always in the direction < 1;-3;-4 > and the tension in the rope is constant at 100 Newtons?

Attempt:

the magnitude of F is 100 Newtons to find its vecor form, do i have to find the unit vector of <1,-3,-4>? r(t) = ti + $t^2$j + $t^3$k and 0 < t < 1?

is that correct?

• You could take a more easier route, hint:as the direction and force of rope is constant, it is just like gravity. – neonpokharkar Sep 20 '17 at 1:51

Your form for $\vec r(t)$ is correct.

To get the vector for $\vec F$, you can write it $$\vec F = F\langle 1,-3,-4\rangle$$ where $F$ is a scalar (that way it goes in the right direction). And then you know its magnitude is $100$ Newtons, so to find $F$ you set $|\vec F| = 100$ and solve.

Then take $\vec F\cdot \vec r'(t)$ and integrate from $t=0$ to $t=1$ to find the work.

• Math Lover deserves the check-mark. I got it in later and more importantly my last line was wrong. – spaceisdarkgreen Sep 20 '17 at 1:50

You do not have to compute the unit vector along $\langle 1,3,-4 \rangle$. All you need is to compute the force vector along this vector. In particular, $$\vec F = c(\mathbf{i}+3\mathbf{j}-4\mathbf{k}),$$ where $|\vec F|=100 \implies c = 100/\sqrt{1^2+3^2+4^2}$. After computing $\vec F$, obtain the work, $W$, by $$W = \int \vec F \vec{\cdot d r}.$$

You could take a more easier route

As the magnitude and direction of the force of constant it is just like gravitational pull.

So you could easily learn from the potential energy equation of gravity $mgh$

For your variable like $h$ , consider the component of displacement in the direction of force.