A constant force of $\overrightarrow{F}=\langle3,7\rangle$ moves an object along a straight line from the point $(-5,-4)$ to the point $(8,9)$. Find the work done if the distances are measured in feet and the force is measured in pounds.
What I Tried
I graphed the two points and found the distance between them. (I did this by drawing a triangle from the straight line created by the points to see what the $X$ and $Y$ portions of the triangle were). But it resulted in the wrong answer.
I used the distance from the two points and multiplied them by $\langle3,7\rangle$ (the force vector). Could you please tell me where I went wrong and what steps I could take to fix this?
Best Answer
The general formula for the work done by a force a path $r$ is:
$$W=F\Delta r cos\theta$$ where $F$ is the magnitude of that force, $\Delta r$ is distance traveled by the particle, and $\theta$ is just the angle between your path and vector that represents the force.
If you've learnt about dot products, you'd notice that this is just the dot product of the force vector and the vector that defines the path between those two points.
What you did wrong there is that you scaled the distance, but not the force. Remember work is a scalar quantity, so multiplying the vector by a distance doesn't work (pun unintended).
As for the "physics" of the matter. What contributes to the work done by a force is only the component of the force in the direction of the movement, $r$, so if you project your force over the line defined between those two points (hence the $cos \theta$), you get the magnitude of the force in the direction of $r$, and then multiplying the two gets you the work done by the force.