What should be the work done here?

[Asad Raza]

Homework Statement

If you apply a force F with the x component =-6.97 N and the y component = 8.47 N on a car which moves along the x-axis by a displacement of 3.37m. What is the work done on the car by your force in J?

Homework Equations

W=Fxd

The Attempt at a Solution

I think that the work done should be zero for this question since the direction of movement is not in the direction of force. Right? Moreover, I was curious because there's no hint in the question that any force acts in positive x direction, albeit the particle moves in that direction.
Kindly answer both parts of the questions

 

[PhanthomJay]

Homework Statement

If you apply a force F with the x component =-6.97 N and the y component = 8.47 N on a car which moves along the x-axis by a displacement of 3.37m. What is the work done on the car by your force in J?

Homework Equations

W=Fxd

The Attempt at a Solution

I think that the work done should be zero for this question since the direction of movement is not in the direction of force. Right?

wrong

Moreover, I was curious because there's no hint in the question that any force acts in positive x direction, albeit the particle moves in that direction.

but there is a force acting in the negative x direction

Kindly answer both parts of the questions

you will have to answer the problem question. What is the definition of work? It is not Fxd.

 

[patrickmoloney]

When you push a car, are you saying no work is done?

The formula for work is W=\vec{F} \cdot s

(hint: Use scalar multiplication)

 

[Asad Raza]

When you push a car, are you saying no work is done?

The formula for work is W=\vec{F} \cdot s

(hint: Use scalar multiplication)

I am asserting no work is done because the force applied and the direction of movement make an angle of 180 degrees with each other. Work done is the distance moved by an object in the direction of applied force.
Now, if I use scalar multiplication, it will render -6.97(3.37)cos(180)=23.4889. Is that you're saying should be he answer?

 

[Asad Raza]

wrong but there is a force acting in the negative x directionyou will have to answer the problem question. What is the definition of work? It is not Fxd.

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?

 

[patrickmoloney]

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?

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

 

[haruspex]

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

No, what you did was a nonsense.
First, displacement is a vector. Work done is $\vec F.\vec s$, the dot representing the dot product, which results in a scalar.

Multiplying a scalar by a vector produces a vector, not the scalar you somehow obtained.
3.37 \begin{bmatrix} -6.97 \\ 8.47 \end{bmatrix}= \begin{bmatrix} 3.37(-6.97)\\ 3.37(8.47)\end{bmatrix}

 

[haruspex]

Work done is the distance moved by an object in the direction of applied force.

Yes, but that is direction in a vectorial sense. No movement in the direction of the force vector would mean that all the movement was at right angles to the force vector.
If I push with force F against an object and move it forward distance s then I have done work Fs. If despite my efforts I go backwards distance s then I have done work F(-s)=-Fs.

 

[RedDelicious]

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?

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.

 

[PhanthomJay]

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?

It doesn't. The car is already moving in the positive x direction before the negative force is applied. The applied negative force slows it down (like a braking force applied to a moving vehicle) as it continues to move in the positive x direction..