Asad Raza said:
That is what I am asking that how can a force in the negative x direction make an object move in the positive x direction?
The force in the -x direction is not what's making it move in the positive x direction. There is another force acting on it. You don't have to worry about it because you're only concerned in finding how much work you've done. Imagine you are lifting weights, doing something like bench press, and you are pushing your hardest to raise the bar but it's too heavy and so it gradually falls back to your chest. As you're pushing against it to raise the bar, you are doing work on the bar, but because it's moving in the opposite direction, you are doing negative work (gravity would be doing positive work)
In this case, the person is pushing against the car but it is still moving in the opposite direction and so the person would be doing what kind of work would they be doing on the car?
patrickmoloney said:
W = \vec{F} \cdot s = s \begin{bmatrix} \vec{x} \\ \vec{y} \end{bmatrix}= 3.37 \begin{bmatrix} -6.97 \\ 8.47 \end{bmatrix}= 3.37(-6.97) + 3.37(8.47) =
If you don't know what I just did, it's simple scalar multiplication. That's what F \cdot s means
This is incorrect. An immediate red flag should have been the fact that you have a contribution from the y-component of the force even though there is no displacement in the y-direction, meaning it should be zero.
Your idea of matrix multiplication is wrong as well. If you were to represent the vectors as matrices, you would be attempting to multiply a (1x1) matrix by a (2x1), which has no defined product. Importantly, matrix multiplication yields another matrix, not a scalar.
In this case,
<br />
<br />
W = \vec{F} \cdot \vec{s} = F_xs_x + F_ys_y<br />
<br />
where the s components are the displacements in the x and y directions respectively.
Note that matrix multiplication multiplies a (1x2) by a (2x1) matrix. The "inner sizes" are both 2 and so the multiplication is valid. For matrix multiplication to be defined, the number of columns on the left must equal the number of rows on the right.