Work done by a force going around a circle ?

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SUMMARY

The discussion focuses on calculating the work done by a force acting on an object moving in a circular path, specifically using the equation Ep(initial) + Ek(initial) + W(Forward) = Ep(Final) + Ek(final) + W(Resistive). The user is attempting to solve for the final velocity (v) and is confused about the term Fs, which represents the resistive work done. The mark scheme indicates that Fs = 0.6 x (40/360) x 2π x 2, where 2π is used to calculate the circumference of the circular path. The key takeaway is understanding how to apply the circular motion concepts to calculate work done.

PREREQUISITES
  • Understanding of potential energy (Ep) and kinetic energy (Ek) in physics
  • Familiarity with the concepts of work and energy conservation
  • Knowledge of circular motion and how to calculate circumference
  • Basic algebra for solving equations involving variables
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  • Study the principles of work-energy theorem in circular motion
  • Learn how to derive the circumference of a circle and its applications in physics
  • Explore examples of calculating work done by forces in circular paths
  • Investigate the relationship between angular displacement and linear distance in circular motion
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Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for examples of energy conservation in circular dynamics.

Tangeton
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I have a problem with question 4 part (iv) which is on here http://www.mei.org.uk/files/papers/m208ju_ail7.pdf

Basically, I used Ep(initial) + Ek(initial) + W(Forward) = Ep(Final) + Ek(final) + W(Resistive) equation, where Ep is the potential energy, Ek the kinetic energy and W is the work done.
This finalized to mgh + 1/2mu2 = 1/2mv2 + Fs (since there is no forward work done and h = 0 at the lowest point)

I worked values of the equation:
mgh = 0.15 x 9.8 x 0.4679
1/2mu2 = 1/2 x 0.15 x 2.52
1/2mv2 = 1/2 x 0.15 x v2 <-- need to find v

Fs is the problem; The mark scheme says that Fs = 0.6 x 40/360 x 2pi x 2

I understand that it must be that 0.6 acts over 40/360 of the circle, and then the distance from A to the lowest point is 2m as I've previously worked out, so this is the s. What I don't understand is 2pi, and whether the 2pi relates to the distance or the force... Could anybody explain how Fs, the resistive work done, can be equal to that and maybe give an equation for working out work done when it acts in circular motion?

Thank you for reading and please if you could help me with my question I would really appreciate it...
 
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Tangeton said:
I have a problem with question 4 part (iv) which is on here http://www.mei.org.uk/files/papers/m208ju_ail7.pdf

Basically, I used Ep(initial) + Ek(initial) + W(Forward) = Ep(Final) + Ek(final) + W(Resistive) equation, where Ep is the potential energy, Ek the kinetic energy and W is the work done.
This finalized to mgh + 1/2mu2 = 1/2mv2 + Fs (since there is no forward work done and h = 0 at the lowest point)

I worked values of the equation:
mgh = 0.15 x 9.8 x 0.4679
1/2mu2 = 1/2 x 0.15 x 2.52
1/2mv2 = 1/2 x 0.15 x v2 <-- need to find v

Fs is the problem; The mark scheme says that Fs = 0.6 x 40/360 x 2pi x 2

I understand that it must be that 0.6 acts over 40/360 of the circle, and then the distance from A to the lowest point is 2m as I've previously worked out, so this is the s.

The distance you want is that traveled by the sphere from B to the lowest point, which is 40/360 of the circumference of a circle of radius 2m.

What I don't understand is 2pi, and whether the 2pi relates to the distance or the force... Could anybody explain how Fs, the resistive work done, can be equal to that and maybe give an equation for working out work done when it acts in circular motion?

Thank you for reading and please if you could help me with my question I would really appreciate it...
 
Can anyone explain 2pi?
 
Tangeton said:
Can anyone explain 2pi?
pasmith explained already:
pasmith said:
The distance you want is that traveled by the sphere from B to the lowest point, which is 40/360 of the circumference of a circle of radius 2m.
Look at the part after "40/360".
 

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