(HW) Vectors - finding work done by forces

Question

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!

Answer 1

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 $W=\vec F\cdot\vec r= \frac{1\cdot 20}{\sqrt{14}} + \frac{2\cdot 30}{\sqrt{14}} + \frac{2\cdot 10}{\sqrt{14}}=\frac{100}{\sqrt{14}}$

Answer 2

Let $a=<2,3,1>$ and $b=<1,2,2>$

Note that $$\cos\theta=\frac{\bar { a }\cdot\bar { b} }{||\bar { a }||\cdot||\bar { b}|| }$$

$$\cos \theta=\frac{2+6+2}{\sqrt{2^2+3^2+1^1}+\sqrt{1^2+2^2+2^2}}$$ $$\cos \theta=\frac{10}{\sqrt{14}+3}$$

Now $$W=F\cdot D\cdot \cos\theta$$ $$W=10\cdot3\cdot\frac{10}{\sqrt{14}+3}$$