AS FMech. Help

OP

Thanks for the reply...and ur saying that the weights parallel component doesn't cause the loss of energy?

Reply 7

OP

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

..

Reply 8

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.

(edited 5 years ago)

Reply 9

OP

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?

Reply 10

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.

Reply 11

Notnek

21

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.

Reply 12

OP

I see, thanks everybody!

Reply 13

OP

so the second bold point was a bit misleading, because that mentioned 'total work done' right?

Reply 14

OP

Whilst in the example, the work done against gravity was not included

Reply 15

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.

Reply 16

OP

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?

Reply 17