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Work by a constant force Show that the work done by a constant force field $\mathbf{F}=a \mathbf{i}+b \mathbf{j}+c \mathbf{k}$ in moving a particle along any path from $A$ to $B$ is $W=\mathbf{F} \cdot \overrightarrow{A B}$

The force $\mathbf{F}$ is conservative because all partial derivatives of $M, N,$ and $P$ are zero. $f(x, y, z)=a x+b y+c z+C ; A=$ $(x a, y a, z a)$ and $B=(x b, y b, z b) .$ Therefore, $\int \mathbf{F} \cdot d \mathbf{r}=$

$f(B)-f(A)=a(x b-x a)+b(y b-y a)+c(z b-z a)=$

$\mathbf{F} \cdot A \hat{B}$

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show that the work done by a constant force f equals ai plus BJ Pussy K. Moving along any path from a to B is w equals force times a dot product of a B. So let's say this is Hey, baby, so first thing we can do is break down. What this equation essentially is so f is a factor, and a B is a scaler or, in other words, from point A to B that tells us the distance between a to be so this is essentially distance and a constant force means the field is conservative. So, knowing this information, let's look back to what the equation is for work in integral form. So work an integral form, is the work equals to the integral off the force field times he dot product of the are. But since we know that it's that there is a constant force, that means it is a conservative field, so we can take out F of the integral and get F times the integral of the are and the integral of D. R is essentially just adding up all that the ours between two points In this case, it's between points a to B And let's say this Anthony and infinitely small line is t. R. So again we're essentially adding up all the d ours, which is actually just a B.