Intergal? work done by a non constant force

[Chuck 86]

Homework Statement

A force F = (4.38 x i + 2.84 y j) N acts on an object as it moves in the x direction from the origin to x = 5.52 m. Calculate the work done on the object by the force.

Homework Equations

intergal of F from 0 to 5.52m

The Attempt at a Solution

not sure if I am suppose to intergrate, or what I am supposed to do to start the process

 

[Donaldos]

$$W=\int\limits_{x=0}^{x=5.52}\vec{F}.{\rm d}\vec{x}$$

where $$\vec{x}=x\vec{i}$$ is the position vector.

 

[Chuck 86]

ok so u intergrate the sum of X and Y right?

 

[Gear300]

Not exactly. By the looks of the problem, the object only moved in the x direction. So there is no net work in the y-direction. You only integrate the x-component.

 

[Chuck 86]

So then it would be the intergal of 4x?

 

[Donaldos]

Yes.

$${\rm d}\vec{x}={\rm d}x \vec{i}$$

so

$$\vec{F}.{\rm d}\vec{x}=\left(4.38x \vec{i}+2.84y \vec{j}\right). \vec{i} {\rm d}x=4.38x{\rm d}x$$

$$W=\int\limits_0^{5.52} 4.38x{\rm d}x$$