Frictional force and Work done on a parabolic path

[Fenixx]

Homework Statement

A sack of flour (mass m) can slide along a parabolic track i the vertical plane defined by y = $$\gamma$$x² The coefficient of (dynamic) friction is $$\mu$$. You move this sack from x = 0 to x = x₁ along the track by pulling it with a horizontal force F (you do this quasi-statically, i.e. the sack is not accelerated).

a) Calculate the work done by force F.
b) Discuss your result for the case $$\mu$$ = 0.

Homework Equations

F_fr=-$$\mu$$mg
W = $$\int$$F dl = F x cos $$\theta$$

The Attempt at a Solution

I know that the horizontal force F = F_fr. If there were no frictional force, the work would simply be the change in gravitational potential from x to x₁, correct? W = mg(y(x₁) - y(x))

However, since there is a frictional force, and it is along a curved path, the work done actually depends on the path taken. I am not sure how to factor this part of the force in Since if F is horizontal, wouldn't amount of force that would affect the bag of flour would change with the path? Does this problem involve some type of line integral? I am not too familiar with this type of problem.

Any pointers in the right direction would be greatly appreciated. Thanks in advance for the help.

 

[anathema]

Hello,

Yes, this problem does involve a line integral. The work done by a force along a curved path can be calculated using the formula W = ∫F·ds, where F is the force vector and ds is the infinitesimal displacement vector along the path. In this case, the force vector is F = Ffr = -μmg, and the displacement vector can be written as ds = dx + dy = dx + \gamma2xdx.

Therefore, the work done by the frictional force along the entire path from x = 0 to x = x1 is given by:

W = ∫F·ds = ∫0x1 (-μmg)(dx + \gamma2xdx) = -μmg∫0x1 (1 + \gamma2x)dx = -μmg[x + \gamma4x2]0x1 = -μmg(x1 + \gamma4x12)

This is the work done by the frictional force. To calculate the total work done by the applied force F, we need to add the work done by F along the same path. Since F is horizontal, the angle between F and ds is always 90°, so the work done by F is simply given by:

W = ∫F·ds = F∫0x1 cos(90°)ds = F∫0x1 ds = F(x1 - 0) = Fx1

Therefore, the total work done by the applied force F is given by:

W = Fx1 - μmg(x1 + \gamma4x12)

Now, let's consider the case where μ = 0. In this case, there is no frictional force acting on the sack of flour, so the work done by the applied force F is simply given by:

W = Fx1

This means that the total work done by F is equal to the work done by F in the presence of friction, minus the work done by the frictional force. This makes sense intuitively, as the frictional force is essentially "stealing" some of the work done by F.

I hope this helps. Let me know if you have any further questions.