Find Work from Force Equation

[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!

 

[rwisz]

I would first clarify the given information and assumptions before proceeding with the solution. Some questions I would ask are:

1. Is the force applied at a constant magnitude or does it vary with distance?
2. Is the block moving in a straight line or on an incline?
3. Is the block experiencing any other forces besides the applied force?

Once these questions are answered, I can then determine the appropriate equation to use for calculating work from force. If the force is constant and the block is moving in a straight line, then the first equation, W = integral(F(x)dx), can be used. However, if the force varies with distance or the block is on an incline, then the second equation, W = integral(Fcos(theta)dx), would be more suitable. If the block is experiencing other forces, then the third equation, W = F(x)delta(x), may be used.

In this case, it seems like the force is constant and the block is moving in a straight line. Therefore, the first equation can be used. However, the angle of 30 degrees above the horizon line should be taken into account when calculating the work. This can be done by using the component of the force in the direction of motion, which is Fcos(theta). So the correct equation to use would be W = integral(Fcos(theta)dx).

Plugging in the given values, we get:
W = integral((x^2-2x)cos(30)dx) from 0 to 2
= (sqrt(3)/2) * integral((x^2-2x)dx) from 0 to 2
= (sqrt(3)/2) * ((x^3)/3 - x^2) from 0 to 2
= (sqrt(3)/2) * ((8/3) - 4)
= (4sqrt(3)/3) - (2sqrt(3))
= (2sqrt(3)/3) Joules

So the work done over the first 2 meters is (2sqrt(3)/3) Joules. It is important to include the angle in the calculation as it affects the direction of the force and therefore, the amount of work done.