## Work done by a variable force

Given a two-dimensional variable force $\overrightarrow{F}(x,y)$ acting on a particle moving along a given curved path or moving along an unknown path, how would you determine the work done by this variable force as the particle moves from point $A$ to point $B$

$\color{blue}{\text{Solution Steps}}$

$W = \int \overrightarrow{F}.\overrightarrow{ds}$ $=\int (f_x(x,y) \hat{i} +$ $f_y(x,y) \hat{j}).$$(dx \hat{i} +$ $dy \hat{j})$ $=\int f_x(x,y) dx +$ $f_y(x,y) dy$

Now depending on the given expressions for $f_x(x,y)$ and $f_y(x,y)$, there are 3 ways we can approach the problem to find the work done

a) If $f_x(x,y) = f(x)$ and $f_y(x,y) = g(y)$, then $W=\int (f_x(x,y) dx +$ $f_y(x,y) dy)$ $=\int_{x_i}^{x_f} f(x) dx +$ $\int_{y_i}^{y_f} g(y) dy$

For example, let’s say that $\overrightarrow{F} =$ $x \hat{i} +$ $y \hat{j}$, then $W=$ $=\int_{x_i}^{x_f} x dx +$ $\int_{y_i}^{y_f} y dy$, which you can easily solve.

Note that if you were given $\overrightarrow{F} =$ $K(\cfrac{x}{(x^2+y^2)^{1/2}} \hat{i} +$ $\cfrac{y}{(x^2+y^2)^{1/2}} \hat{j})$ and it was also given that particle is moving along a circular path of radius ‘$a$’, centered at origin, then recall that $(x^2+y^2)^{1/2} = a$ and so, we can write $\overrightarrow{F} =$ $\cfrac{K}{a}(x \hat{i} +$ $y \hat{j})$ and the rest of the steps for finding $W$ will be same as mentioned above.

b) If $f_x(x,y)$ and $f_y(x,y)$ are such that $f_x(x,y) dx + f_y(x,y)dy$ can be expressed as exact differential, i.e. $f_x(x,y) dx + f_y(x,y)dy$ $=d(g(x,y))$ then $W=\int f_x(x,y) dx +$ $f_y(x,y) dy$ $=\int_{x_i, \ y_i}^{x_f, \ y_f} d(g(x,y))$ $=g(x_f,y_f) – g(x_i,y_i)$

Recall that $\cfrac{d(AB)}{dt} =A \cfrac{d(B)}{dt} + B\cfrac{d(A)}{dt}$
Or, $d(AB) = Ad(B) + Bd(A)$

For example, let’s say that $\overrightarrow{F} =$ $xy^2 \hat{i} +$ $x^2y \hat{j}$, then $W=\int (xy^2 dx +$ $x^2y dy)$ $=\int_{x_i, \ y_i}^{x_f, \ y_f} d(\cfrac{1}{2}x^2y^2)$ $=\cfrac{1}{2}(x_f^2y_f^2 – x_i^2y_i^2)$

c) If $f_x(x,y)$ and $f_y(x,y)$ are such that we CANNOT express $f_x(x,y) dx + f_y(x,y)dy$ as exact differential and if the trajectory of the path is specified, say for example $(x-a)^2+(y-b)^2 = c^2$

Then, to solve $W=\int f_x(x,y) dx +$ $f_y(x,y) dy$, we will write $y$ in terms of $x$ using the equation of trajectory for $f_x$ and write $x$ in terms of $y$ for $f_y$, then we should be able to integrate the two terms. Here you will need to be careful when writing $x$ in terms of $y$ or vice versa for the given trajectory as the expression might be different for different parts of the trajectory, so break the integral limits as appropriate and then integrate.

## Work done by a variable force | Sample Problems

Q1 Given $\overrightarrow{F}=$ $xy^2 \hat {i} + x^2y \hat{j}$ newtons, find the work done by $\overrightarrow{F}$ when a particle is taken along the semicircular path $AB$ where the coordinates of $B$ are ($2,0$)

Zero.

$W = \int F_x dx + F_y dy$ $=\int (xy^2 dx +$ $x^2y dy)$ $=\int_{x_i, \ y_i}^{x_f, \ y_f} d(\cfrac{1}{2}x^2y^2)$ $=\cfrac{1}{2}(x_f^2y_f^2 – x_i^2y_i^2) = \cfrac{1}{2}(4.0-0.0) = 0$

Q2. Particle moves from point A to point B along the line shown in figure under the action of force $\overrightarrow{F}$ $=−x \hat{i} + y \hat{j}$. Determine the work done on the particle by $\overrightarrow{F}$ in moving the particle from point $A$ to point $B$ (all quantities are in SI units).

1J

$W = \int F_x dx + F_y dy$ $=\int (-x dx +$ $y dy)$ $=-\int_{1}^{0} xdx$ $+\int_{0}^{1} ydy$