A Force Changing With Distance (bead moving on a curved wire)

In summary, there is a problem involving an equation that requires the utilization of energy and a changing acceleration. The equation for energy conservation is also mentioned, as well as a force that linearly increases with time. However, there are some clarifications needed, such as a diagram, the shape of the curved wire, and the initial velocity. Additionally, it is stated that the only force acting on the system is gravity, but this seems to contradict the description of a force increasing linearly with distance. Further clarification is needed before a solution can be provided.
  • #1
gkg
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0
Homework Statement
A bead moving along a curved wire is acted upon by a downwards force that increases linearly with distance from the initial x-position. The bead arrives at the end of the wire a distance of L from its initial position in t amount of time. Derive an expression to find the final velocity of the bead. Assume that the wire is frictionless and the only force acting upon the system is gravity.
Relevant Equations
v(f) = v(0) + a*t
I initially tried to solve this equation using work, but was stuck in a confusing integral that didn't make sense. I am almost sure that the utilization of energy is needed to solve this equation, but I have been flustered for the past three days at solving this.
 
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  • #2
That formula you wrote assumes constant acceleration.

Why not post what your actual attempt was, i.e. that integral you mentioned.

How would energy conservation equation look like? Post it and I can give you a hint how to proceed
 
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  • #3
With a changing acceleration that equation would be:
v(f) = v(0) + (da/dt) t^2

Assuming a force that linearly increases with time, the work equation would be:
W = Integral(m*a*x) dl, where max is the force equation and dl is the line integral.

Since energy is not actually conserved and it seems to be increasing due to the increasing force the energy equation would be:

1/2*m*v(f)^2 + U(f) = U(0) +1/2*m*v(0)^2
Where U stands for the potential energy, as the bead does go up in the y-direction from the curve of the wire that it is on.
 
  • #4
gkg said:
With a changing acceleration that equation would be
You can see for yourself if this is not correct. Hint, change in v is the integral of a.

Do you know of a real force that grows with distance? Learned about Hookes law? Isnt such force conservative?
 
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  • #5
gkg said:
v(f) = v(0) + (da/dt) t^2
How did ##\int a\cdot dt## become ##da/dt\cdot t^2##?
gkg said:
Assuming a force that linearly increases with time,
On what basis? You are told it increases linearly with distance.
 
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  • #6
Hi @gkg and welcome to PF.

In addition to the other replies, please note that are some problems with the question.

Is the question complete? The issues that spring to mind are:
- Is there a diagram?
- Is the shape (or at least the positions of the start and end points) of the curved wire defined ?
- Is the initial velocity (or maybe that should be speed) zero? Or some given quantity?
- The question asks for the final velocity; this would require a direction as well as a magnitude – should the question be asking for final speed?

Also, you may find it helpful to read the ‘rules’ so you know what to expect here! https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/
 
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  • #7
If the only force is gravity, what is the point of describing that force which increases linearly with distance? How is this force relevant to the problem? Or is gravity supposed to be this force increasing linearly with distance? The problem needs clarification before asking for a solution. And what Steve4Physics wrote, too.
 
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  • #8
nasu said:
If the only force is gravity, what is the point of describing that force which increases linearly with distance? How is tReplyhis force relevant to the problem?
In fact the question says "... the only force acting upon the system [my underlining] is gravity.".

Assuming that's a correct staement, it implies the system is in free-fall. That's probably not the intention!

We await clarification from @gkg.
 
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