# AS FMech. HelpWatch

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#1
Hi all, can someone pls help me understand the example I'm about to link up
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#2
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#3
Why haven't they accounted for the component of weight acting down the slope along with friction as the "total" work done?
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#4
Bump
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#5
Any clarification wud be much appreciated
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11 months ago
#6
The loss in Energy is the work done due to friction, it's the same formula rearranged.

If there was no friction ∆KE=∆GPE, but because there is ∆KE = ∆GPE + work done due to frictional forces
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#7
Thanks for the reply...and ur saying that the weights parallel component doesn't cause the loss of energy?
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#8
(Original post by Luqman.ldn)
The loss in Energy is the work done due to friction, it's the same formula rearranged.

If there was no friction ∆KE=∆GPE, but because there is ∆KE = ∆GPE + work done due to frictional forces
..
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11 months ago
#9
I'd presume it's taken into account because the velocity of the particle is decreasing. We are gaining GPE because it has velocity but that velocity is decreasing due to the component of gravity acting on it.
As velocity decreases kinetic energy also decreases, so yes it does cause a loss in energy.
Last edited by Luqman.ldn; 11 months ago
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#10
(Original post by Luqman.ldn)
I'd presume it's taken into account because the velocity of the particle is decreasing. We are gaining GPE because it has velocity but that velocity is decreasing due to the component of gravity acting on it.
As velocity decreases kinetic energy also decreases, so yes it does cause a loss in energy.
So then surely should be factored in along with friction equating to the loss in energy rit?
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11 months ago
#11
(Original post by Makise kurisu)
So then surely should be factored in along with friction equating to the loss in energy rit?
We have, the loss in kinetic energy.
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11 months ago
#12
(Original post by Makise kurisu)
Thanks for the reply...and ur saying that the weights parallel component doesn't cause the loss of energy?
This is what GPE is. Notice the expression they used for GPE:

2 x 9.8 x xsin(45)

If you found work done against gravity then you would get the same thing.
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#13
I see, thanks everybody!
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#14
so the second bold point was a bit misleading, because that mentioned 'total work done' right?
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#15
Whilst in the example, the work done against gravity was not included
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11 months ago
#16
The work done against gravity is included, it's MGH.
mg is the force and h is the distance moved in the direction of the force.
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#17
Hmm...but now if u read the text between the bold points it implies that gravity has no affect on the total energy change of an object, doesn't that contradict with what the above posts establish?
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11 months ago
#18
(Original post by Makise kurisu)
Hmm...but now if u read the text between the bold points it implies that gravity has no affect on the total energy change of an object, doesn't that contradict with what the above posts establish?
gravity cannot affect the total energy change, an external force can. Because if ∆KE=∆GPE and total energy= ∆KE ∆GPE, if GPE does change Ur total energy doesn't change, but when there is another external force acting, it's the only thing affecting the total energy.
Last edited by Luqman.ldn; 11 months ago
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