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

AI Thread Summary
To calculate work from a force applied at an angle, the appropriate equation is Work = integral(Fcos(theta)dx). The force is given as F(x) = x^2 - 2x, and the angle is 30 degrees, leading to the cosine component being cos(30) = sqrt(3)/2. By integrating the expression sqrt(3)/2(x^2 - 2x) over the distance of 2 meters, the result is found to be -2sqrt(3)/3. This confirms the correct application of the work formula in this scenario.
gggorillaz
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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?
 
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gggorillaz said:

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.
 
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?
 
gggorillaz said:
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!:approve:
 
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