I have the following question:
A force $F$ with a magnitude of $10$ newtons has the same direction (but different magnitude) as the vector $2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$. This force pushes an object on a ramp in a straight line from the point $(3, 1, 5)$ to the point $(4, 3, 7)$, where the coordinates are measured in
meters. How much work is done by the force?
I want to make sure I am doing this correctly or on the right track.
I know $\mathrm{work} = \mathrm{force} \times \mathrm{distance}$, and the distance between the two points is $3\mathrm{m}$ by the distance formula. I am confused on what to do next.
Is $\mathrm{force} = 10(2\mathbf{i} + 3\mathbf{j} + \mathbf{k})$? I feel like I am missing something simple and would like a hint. Thanks!
Best Answer
Directions are defined by unit vectors. Since we know $\vec F$ is in the direction of $2\hat i + 3\hat j + \hat k$ which has magnitude $\sqrt{14}$ and the magnitude of $\vec F$ is $10$, $\vec F = \frac{10}{\sqrt{14}}(2\hat i + 3\hat j + \hat k)= \frac{20}{\sqrt{14}}\hat i + \frac{30}{\sqrt{14}}\hat j + \frac{10}{\sqrt{14}}\hat k$.
We can get the distance vector $\vec r$ by taking the difference between the ending and starting points to get $\vec r=(4-3)\hat i + (3-1)\hat j + (7-5)\hat k=\hat i + 2\hat j + 2\hat k$.
Finally we take the dot product to get $W=\vec F\cdot\vec r= \frac{1\cdot 20}{\sqrt{14}} + \frac{2\cdot 30}{\sqrt{14}} + \frac{2\cdot 10}{\sqrt{14}}=\frac{100}{\sqrt{14}}$