Question

Question posted below...

Answer 1

A force...
\[\vec F = (4x \vec i + 3y \vec j)\]Where F is in newtons, and x and y are in meters, acts on an object as the object moves in the x direction from the origin to x = 5.00m. Find the work...\[W=\int\limits_{}^{}\vec F \cdot d \vec r\]Done by the force on the object.

Answer 2

Let me give this a shot, and see if I'm doing it right so far...

Would it be...\[W=d \vec r \cdot \int\limits_{0m}^{5.00m} \vec F\]\[W=d \vec r \cdot \int\limits_{0m \vec i}^{5.00m \vec i}(4x \vec i + 3y \vec j)=d \vec r \cdot \int\limits_{0m \vec i}^{5.00m \vec i}(4x \vec i + 3 y \vec j)\]

Answer 3

If what I've done above is correct, I'm pretty lost past that. If anyone could help?

Answer 4

You are moving from the origin to (5,0). So dr is always in the x direction; in fact write it as
\[ \vector{dr} = \hat{i} dx \]

Answer 5

Hence
\[ F \cdot dr = (4x \ i + 3y \ j ) \cdot i \ dx = 4x \ dx \] That is to say, the component of F that is perpendicular to the direction of motion contributes no work whatsoever.

Make sense?

Answer 6

Yes.

So...

\[W=\int\limits_{0m}^{5.00m} \vec F \cdot d \vec r = \int\limits_{0m}^{5.00m}4x\ dx = 2(5)^2 - 2(0)^2=50.0J\]?

Answer 7

Yes

Answer 8

Ok. Thanks, James!

I may have another question related to work here in a bit.

Answer 9

ok. Post it as a new question and try and get my attention if I'm around.