How to calculate work with i and j?

Question

I don't understnad how to solve. The problem is:

w = ( 2.0 N i - 0.6 N j ) X (-3 m i)

Is it supposed to be: (2.0N)(-3m) i - 0.6N J ?

Answer 1

What the book asked you to calculate is apparently $$ W = (2.0\mathrm N \hat\imath - 0.6\mathrm N \hat\jmath) \cdot (-3\mathrm m\hat\imath). $$

What you calculated is $$ \big((2.0\mathrm N \hat\imath) \cdot (-3\mathrm m\hat\imath)\big) - (0.6\mathrm N \hat\jmath). $$

These are not the same thing. An analogy would be if you were asked for $(8 - 5) \cdot 3$ and you calculated $(8 \cdot 3) - 5$ instead; in that case it is easy to work out the two results and see that they are different.

There is a shortcut we can take for the "dot" product in which we multiply only the terms that have matching unit vectors--that is, we multiply the $\hat\imath$ terms with each other and the $\hat\jmath$ terms with each other. But that shortcut only works if you multiply all terms, $\hat\jmath$ as well as $\hat\imath.$

Also, the "dot" product of two vectors is just a number. For example, $2.0\hat\imath \cdot 3.0\hat\imath = 6.0.$ There should be no $\hat\imath$ or $\hat\jmath$ in your answer.

The fact that there doesn't appear to be a $\hat\jmath$ term in the "distance" vector in the question does not allow you to merely copy the $\hat\jmath$ term that you happen to see in the "force" vector. It means that in order to apply the shortcut, you must write out both terms of the "distance" vector, either on paper or in your head, before you multiply. That is, you can write the formula like this: $$ W = (2.0\mathrm N \hat\imath - 0.6\mathrm N \hat\jmath) \cdot (-3\mathrm m\hat\imath + 0\mathrm m \hat\jmath). $$

So from the $\hat\imath$ terms we get the product $(2.0\mathrm N)(-3\mathrm m) = -6\mathrm J,$ because Newtons multiplied by meters are equal to Joules. (The J in the book's answer is a symbol for the units, Joules, and has nothing to do with the vector $\hat\jmath.$) For the $\hat\jmath$ term we have zero, because one of the terms is $0\mathrm m \hat\jmath$ and anything times zero is zero. So the final answer is $6\mathrm J + 0\mathrm J = 6\mathrm J.$

---

By the way, one quick way to see that your original answer is wrong is to look at the units. Work has units of Newton-meters (Nm) or Joules (J), and it doesn't have a direction. Your proposed answer, $-6 \mathrm N\mathrm m \hat\imath - 0.6\mathrm N \hat\jmath,$ is an attempt to subtract something from something with different units; it makes as much (or as little) sense as trying to subtract $0.6$ meters from $6$ square meters, or $0.6$ meters from $6$ meters per second. Your answer also is some kind of vector (because its components are multiples of $\hat\imath$ and $\hat\jmath$) when it should just be a number of Joules.