Work done by a force going around a circle ?

In summary, The conversation is about a problem with question 4 part (iv) from a physics paper. The equation used is Ep(initial) + Ek(initial) + W(Forward) = Ep(Final) + Ek(final) + W(Resistive), with Ep representing potential energy, Ek representing kinetic energy, and W representing work done. The values for each variable are calculated and the problem arises with finding Fs, the resistive work done. The mark scheme states that Fs is equal to 0.6 x 40/360 x 2pi x 2, with 2pi representing the distance traveled by the sphere from point B to the lowest point. The conversation concludes with the asker requesting an explanation for 2
  • #1
Tangeton
62
0
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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  • #2
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...
 
  • #3
Can anyone explain 2pi?
 
  • #4
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".
 

What is work done by a force going around a circle?

The work done by a force going around a circle refers to the amount of energy transferred to an object as it moves in a circular path. It is a measure of the force applied to the object and the distance it travels along the circular path.

How is work done by a force going around a circle calculated?

The work done by a force going around a circle is calculated using the equation W = F x d x cosθ, where W is the work done, F is the force applied, d is the distance along the circular path, and θ is the angle between the force and the direction of motion.

What is the unit of measurement for work done by a force going around a circle?

The unit of measurement for work done by a force going around a circle is joules (J), which is equivalent to Newton-meters (N∙m).

What happens to the work done by a force going around a circle if the force is perpendicular to the direction of motion?

If the force is perpendicular to the direction of motion, the work done is equal to zero, as the cosine of 90 degrees (perpendicular angle) is equal to 0. This means that the force does not transfer any energy to the object, and thus, no work is done.

How does the radius of the circle affect the work done by a force going around a circle?

The radius of the circle does not affect the work done by a force going around a circle. As long as the force remains constant and perpendicular to the direction of motion, the work done will remain the same regardless of the radius of the circle.

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