How Do You Calculate Work from a Force Applied at an Angle?

[gggorillaz]

Homework Statement

A force is applied to a block downwards at 30 degrees above the horizon line. the Force = x^2 - 2x (Newtons). Find work over the first 2 meters

Homework Equations

Work = integral(F(x)dx)
Work = integral(Fcos(theta)dx)
Work = F(x)delta(x)

The Attempt at a Solution

Im really unsure as to which equation i should use in this case. If I use the top one i find that (x^3)/3 - x^2 = W so I would plug in 2 for x and get 8/3 - 4 but that leaves out angle so I think it's incorrect.
The middle equation looks more promising but I am confused as to what F is in this case. would F be the F(x)? if so would i have to integrate (x^2 - 2x)*sqrt(3)/2 (cos30 = sqrt(3)/2)
The lower equation looks just as wrong as the first, so I am leaning towards the middle one to help me out. Is this even remotely correct?

 

[PhanthomJay]

Homework Statement

A force is applied to a block downwards at 30 degrees above the horizon line. the Force = x^2 - 2x (Newtons). Find work over the first 2 meters

Homework Equations

Work = integral(F(x)dx)
Work = integral(Fcos(theta)dx)
Work = F(x)delta(x)

The Attempt at a Solution

Im really unsure as to which equation i should use in this case. If I use the top one i find that (x^3)/3 - x^2 = W so I would plug in 2 for x and get 8/3 - 4 but that leaves out angle so I think it's incorrect.
The middle equation looks more promising but I am confused as to what F is in this case. would F be the F(x)? if so would i have to integrate (x^2 - 2x)*sqrt(3)/2 (cos30 = sqrt(3)/2)
The lower equation looks just as wrong as the first, so I am leaning towards the middle one to help me out. Is this even remotely correct?

Yes, it is. The middle equation is the definition of work.

 

[gggorillaz]

Ok sweet, so i would find integral of sqrt(3)(x^2/2 -2x) which = sqrt(3)/2(x^3/3 - x^2). now if i plug 2m in for x i find sqrt3 /2(8/3 - 4) which =-2sqrt(3)/3 right?

 

[PhanthomJay]

Ok sweet, so i would find integral of sqrt(3)(x^2/2 -2x) which = sqrt(3)/2(x^3/3 - x^2). now if i plug 2m in for x i find sqrt3 /2(8/3 - 4) which =-2sqrt(3)/3 right?

Yes, correct!