# Work Done By Force Field on Particle (Vector Fields) Ex2

##### Work Done By Force Field on Particle (Vector Fields, Two Points) Example 2

Find the work, W, done by the force field F in moving an object from point P to Q, where P is (1,1) and Q is (2,4).

We’re given $$F(x,y) = 2{y^{\frac{3}{2}}}\mathop i\limits^ \to + 3x\sqrt y \mathop j\limits^ \to$$.

First, even if you aren’t told this, you should check and ask “Is F a conservative vector field?”

If it isn’t, the problem can’t be solved.

Let’s check to see if F is a conservative vector field.

To be conservative for two variables, the partial derivative for the first component of the vector field with respect to x must equal the partial derivative for the second component of the vector field with respect to y.

This is a mouthful, but basically derive what’s before i with x, and derive what’s before j with y and check if equal.

Let Q represent the i component and P represent the j component.

$F(x,y) = 2{y^{\frac{3}{2}}}\mathop i\limits^ \to + 3x\sqrt y \mathop j\limits^ \to$

$\frac{{\partial Q}}{{\partial x}} = 3\sqrt y$

$\frac{{\partial P}}{{\partial y}} = 3\sqrt y$

$\frac{{\partial Q}}{{\partial x}} = \frac{{\partial P}}{{\partial y}}$

So yes, the vector field given is indeed a conservative vector field.

We can find the work from the following integral.

$W = \int {F \cdot dr}$

We need f, which is a function that, when you find the gradient of it, equals F.

You can sometimes do this by “eyeballing it”, but a more systematic way is to integrate with respect to x, then derive that with respect to y.

$f = \int {2{y^{\frac{3}{2}}}} dx$

$f = 2x{y^{\frac{3}{2}}} + g(y)$

Where g(y) is an unknown function of y.

Now derive this f with respect to y.

$\frac{{\partial f}}{{\partial y}} = 3x\sqrt y + g'(y)$

This needs to be equal to what is in the j component of the vector field given.

It’s clear that g'(y) = 0, and integrating that, it’s clear that g(y) = K, where K is just a number constant.

So,

$f(x,y) = 2x{y^{\frac{3}{2}}} + K$

Did you get f correct?

To check, just find the gradient of f and you need to be able to get back the vector field given in the question.

$\nabla f = \left\langle {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}}} \right\rangle$

$\nabla f = \left\langle {2{y^{\frac{3}{2}}},3x\sqrt y } \right\rangle$

So our f is good!

Remember, you can write out the components with i, j, k, or you can just put it in a vector with brackets as shown above.

For finding work, just substitute the points into f and find the difference between the endpoint and the first point.

$W = \int {F \cdot dr}$

$W = f(2,4) – f(1,1)$

$W = \left( 2 \right)\left( 2 \right){\left( 4 \right)^{\frac{3}{2}}} + K – \left( 2 \right)\left( 1 \right){\left( 1 \right)^{\frac{3}{2}}} – K$

$$W = 30$$ $$N \cdot m$$

So the work done is 30 Nm.

We weren’t given units in the question, but in general work has units of Joules (J), which is just Newton-meters $$(N \cdot m)$$.

Return to the previous example, work done by vector field onto particle