Homework Help: Calculate the work done by a force on a particle

1. Feb 3, 2016

henrco

1. The problem statement, all variables and given/known data
A force Fx acts on a particle that has a mass of 1.46 kg. The force is related to the position x of the particle by the formula Fx = Cx^3, where C = 0.550 if x is in meters and Fx is in newtons. Calculate the work done by this force on the particle as the particle moves from x = 2.96 m to x = 1.47 m.

2. Relevant equations
W = F.x
Where F is the force and x is the displacement

3. The attempt at a solution
The displacement x = 1.47-2.96 = -1.49m
The force Fx = Cx^3 = (0.550)(-1.49)^3 = -1.82 (to 3 significant figures)
Input these values into the formula, W=F.x

W = (-1.82)(-1.49) = 2.71 Joules (to 3 significant figures)

This is the answer I get however it's not the expected answer. Could someone please point me in the right direction, so I can understand where I am going wrong.

2. Feb 3, 2016

cnh1995

This is true when the force is constant. What is the fundamental equation of work in terms of force and displacement? You need integral calculus for that.

3. Feb 3, 2016

PeroK

Here's how I interpret that equation:

The work done, $W$, by a constant force $F$, acting over a distance $x$ equals $Fx$

Is that a relevant formula in this case?

4. Feb 3, 2016

ChrisVer

the work can be the area between your $F(x)$ and the x-axis from $x= x_{in}$ to $x=x_{fin}$ ...

5. Feb 3, 2016

henrco

Thank you gentlemen, that was very helpful. Here's a second attempt.

So because it's a variable force I need to integrate the area under the curve.

W = ∫ F(x) . dx (from x = 2.96 m to x = 1.47 m). Where F(x) and dx are both vectors - unable to write vector notation.

The dot product of two vectors:

F(x) . dx = F(x).dx.Cos(180). It's 180 as it's moving in the opposite direction. Is this correct?
F(x).dx.(-1)
-(F(x).dx)

Inputing the value for F(x) into the equation.
W = ∫ -(Cx^3) dx (from x = 2.96 m to x = 1.47 m)
W = - Cx^3 x (from x = 2.96 m to x = 1.47 m)
W = - C(1.47)^3(1.47) - (-C(2.96)^3(2.96))
W = - ((0.550)(1.47)^3)(1.47) - ((-0.550)(2.96)^3)(2.96)
W = -2.57 +42.2
W = 39.7 Joules

6. Feb 3, 2016

PeroK

That's one way to look at it. Another is that $\int_a^b = -\int_b^a$

Something's gone wrong with your integration there. First, you already reversed the integral value by taking the negative, but then you left the end points going high to low. It's one or the other. Second, you didn't integrate $x^3$ correctly.[/QUOTE]

7. Feb 4, 2016

henrco

PeroK, thank you for that.

I have now reversed the range, so it is from a= 1.47 to b= 2.96.
Also hopefully I've correctly integrated x^3.

W = - ∫ (Cx^3) dx (from a =1.47 to b = 2.96)
W = - ((C/4) x^4) + C (from a =1.47 to b = 2.96)

(I'm little confused by the constant of integration and the constant C provided in the question. I've discarded the constant of integration as I would normally when inputting a range).

W = - ((0.550)/4) (2.96)^4) - ((0.550)/4) (1.47)^4)
W = - 10.6 + 0.642
W = -9.96 Joules

8. Feb 4, 2016

ChrisVer

For definite integrals, there is no need for the constant of integration... it dies out after the subtraction.
You got confused because you used the same symbol for both. Either use different symbols or even if you use the same, avoid them having the same capitalization/font etc.

9. Feb 4, 2016

ChrisVer

are you sure you did the calculations right?

10. Feb 4, 2016

henrco

Thanks for feedback and for checking the calculations ChrisVer,

I redid the calculations more carefully and got.

W= -9.91 Joules

11. Feb 5, 2016

henrco

Hi,

Would anyone please check my last post, to see if I've calculated the final result correctly.
ChrisVer said I had made a mistake but I'm not sure if I've rectified it.

Thanks.

12. Feb 5, 2016

cnh1995

Looks correct to me!

13. Feb 5, 2016

henrco

I input the answer -9.91 Joules as the submission was electronic and it says my answer was wrong!!!

Could someone else please have a look at my calculation and let me know what I've done wrong.

14. Feb 5, 2016