How to Calculate Work on an Object with Varying Position-dependent Force?

In summary, the object experiences a force in the x-direction whose magnitude depends on the position of the object, with alpha (not zero). The work done on the object is calculated by multiplying the force by the displacement.
  • #1
mcmuffinpopper
3
0

Homework Statement



An object has several forces acting on it. One of these forces is F=alpha xy ihat, a force in the x-direction whose magnitude depends on the position of the object, with alpha (not zero). Calculate the work done on the object by this force for the following displacements of the object.

1.The object starts at point x=x0, y=0 and moves in the y-direction to the point x=x0 and y=y1

2. The object starts at the origin (x,y)=(0,0) and moves on the line y = 1.5x to the point x = x2, y = y2.

Homework Equations


W = F*s

The Attempt at a Solution


I don't really know how to solve this as I just started with vectors and I still have problems understanding them. I assumed for the 1st one that there was no work done because x stays zero as it is moving parallell to the x-axis while staying 0, so F*s= 0. I don't know how to solve the second one.
 
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  • #2
mcmuffinpopper said:
1.The object starts at the point x = 0, and moves in the y-direction to the point x = x_0 and y = y_1.
Are you sure you have stated that correctly? x=0 is not a point. If it starts at point where x=0 and moves in the y direction then, as you say, x will still be 0, so it seems odd to say it finishes at (x0,y1). But if that is what it does then yes, the force is always zero so the work is zero.

For the second part, you need to write your W=Fs equation in vectors, and allowing for the displacement vector to be infinitesimal. Do you know what I mean by that, and how to do it?
 
  • #3
haruspex said:
Are you sure you have stated that correctly? x=0 is not a point. If it starts at point where x=0 and moves in the y direction then, as you say, x will still be 0, so it seems odd to say it finishes at (x0,y1). But if that is what it does then yes, the force is always zero so the work is zero.

For the second part, you need to write your W=Fs equation in vectors, and allowing for the displacement vector to be infinitesimal. Do you know what I mean by that, and how to do it?
I'm sorry I have indeed stated that incorrectly. It is:
1. The object starts at point x=x0, y=0 and moves in the y-direction to the point x=x0 and y=y1

And I think you mean that I need to integrate W=F*s? I know that if you want to write it in vectors you should write it with arrows on top of them, I don't really know how I should proceed further, so no I don't know how to do it.
 
  • #4
mcmuffinpopper said:
I'm sorry I have indeed stated that incorrectly. It is:
1. The object starts at point x=x0, y=0 and moves in the y-direction to the point x=x0 and y=y1

And I think you mean that I need to integrate W=F*s? I know that if you want to write it in vectors you should write it with arrows on top of them, I don't really know how I should proceed further, so no I don't know how to do it.
Ok, the vectorial calculus form is ##\int \vec F.\vec {ds}## where the dot is the dot product of the vectors. ##\vec {ds}## is the vector (dx,dy), or in the hat notation ##\hat i dx+\hat j dy##. Plug in the vector form of the force you are given, and see where that takes you.
 
  • #5
Is this correct?

J9rvgem.jpg
 
  • #6
That image took ages to load and came out sideways. Parts are also hard ro read. Please take the trouble to type in your algebra.
It was ok down to where you wrote out the force as a vector in hat notation You need to do the same for ##\vec {ds}## and execute the dot product.
Edit: if you cannot write LaTeX, don't worry about the arrows and hats.
It's late here... off to bed.
 

1. How do you calculate work on an object?

Work on an object can be calculated by multiplying the force applied to the object by the displacement of the object in the direction of the force. The formula for work is W = F * d.

2. What is the unit of measurement for work?

The unit of measurement for work is joules (J), which is equivalent to a newton-meter (N*m).

3. Does the direction of the force affect the work done on an object?

Yes, the direction of the force does affect the work done on an object. Work is only considered to be done when the force and displacement are in the same direction. If the force and displacement are in opposite directions, no work is done on the object.

4. Can negative work be done on an object?

Yes, negative work can be done on an object. This occurs when the force and displacement are in opposite directions, resulting in a negative value for work. Negative work can also be interpreted as work being done against the direction of motion of the object.

5. What are some examples of work being done on an object?

Examples of work being done on an object include lifting a book, pushing a shopping cart, and pulling a suitcase. Any time a force is applied to an object and it moves in the direction of the force, work is being done on the object.

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