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# M2 Work, Energy & Power Question watch

1. I'm not sure what I'm doing wrong at part ii)

I considered the system as a whole from A-D.
Work done against resistance from B-C = 14x25 = 350J
Change in GPE= mgh = 10 x 9.8 x 7 = 686J
Change in KE = 0.5mv^2-0.5mu^2 = 0.5x10xv^2-0.5x10x16.6^2 = 5v^2-1377.8

Work done = energy lost - energy gained
350 = (5v^2 - 1377.8) - 686
But this results in a value of v as 21.972m/s

Not sure where I'm going wrong as the answer should be 8.27m/s :/

Any help would be great! Thanks!
2. (Original post by F3rnw3h)
I considered the system as a whole from A-D.
Work done against resistance from B-C = 14x25 = 350J
Change in GPE= mgh = 10 x 9.8 x 7 = 686J
Change in KE = 0.5mv^2-0.5mu^2 = 0.5x10xv^2-0.5x10x16.6^2 = 5v^2-1377.8

Work done = energy lost - energy gained
350 = (5v^2 - 1377.8) - 686
But this results in a value of v as 21.972m/s

Not sure where I'm going wrong as the answer should be 8.27m/s :/

Any help would be great! Thanks!
5v^2-1377.8 is the negative of the loss in energy.
The loss in energy is initial - final.
3. (Original post by ghostwalker)
5v^2-1377.8 is the negative of the loss in energy.
The loss in energy is initial - final.
Thankyou! So using the equaion
work done = energy lost - energy gained
if it had been GPE that was lost, 'energy lost' would be the negative of mgh?
4. (Original post by F3rnw3h)
Thankyou! So using the equaion
work done = energy lost - energy gained
if it had been GPE that was lost, 'energy lost' would be the negative of mgh?
Energy loss is a positive quantity. If h was negative then it would be -mgh, yes.

I find the use of loss and gain can lead to confusion, and always go back to a more basic equation.

Initial energy + energy input = final energy + work down.

Energy input is usually zero, so equation simplifies to

Initial energy = final energy + work down.

Use that as the basis, and rearrange if required.
5. (Original post by ghostwalker)
Energy loss is a positive quantity. If h was negative then it would be -mgh, yes.

I find the use of loss and gain can lead to confusion, and always go back to a more basic equation.

Initial energy + energy input = final energy + work down.

Energy input is usually zero, so equation simplifies to

Initial energy = final energy + work down.

Use that as the basis, and rearrange if required.
Thankyou! How would you go about using that equation with this question though
6. (Original post by F3rnw3h)
Thankyou! How would you go about using that equation with this question though
Initial KE + initial GPE =final KE + final GPE + work done.

Rearrange, if desired:

(Initial KE - final KE) + (initial GPE - final GPE) = work done.

Using initial and final, eliminates the necessicty of considering whether we have a loss or gain.
7. (Original post by ghostwalker)
Energy loss is a positive quantity. If h was negative then it would be -mgh, yes.

I find the use of loss and gain can lead to confusion, and always go back to a more basic equation.

Initial energy + energy input = final energy + work down.

Energy input is usually zero, so equation simplifies to

Initial energy = final energy + work down.

Use that as the basis, and rearrange if required.
I find that students often get confused when considering gain/loss in energy. But it's the most commonly taught method and the one you see in textbooks so I often have to assume students use this method.

I agree that for the average student it is probably better to use a bigger formula (slightly rearranged version of yours):

Initial energy + work in - work out = final energy
8. (Original post by ghostwalker)
Initial KE + initial GPE =final KE + final GPE + work done.

Rearrange, if desired:

(Initial KE - final KE) + (initial GPE - final GPE) = work done.

Using initial and final, eliminates the necessicty of considering whether we have a loss or gain.
So would initial GPE in this question be zero?
9. (Original post by Notnek)
I find that students often get confused when considering gain/loss in energy. But it's the most commonly taught method and the one you see in textbooks so I often have to assume students use this method.

I agree that for the average student it is probably better to use a bigger formula (slightly rearranged version of yours):

Initial energy + work in - work out = final energy
Thankyou! Confused what would be meant by ‘work in’ and ‘work out’ in the context of the question I posted above though
10. (Original post by F3rnw3h)
Thankyou! Confused what would be meant by ‘work in’ and ‘work out’ in the context of the question I posted above though
I shouldn't have disrupted the thread with a different formula (even though they're basically the same). Try doing the question using Ghostwalker's method
11. (Original post by F3rnw3h)
So would initial GPE in this question be zero?
Yes.

You can assign any point as the zero height mark, since it's the difference in the GPE's that's important, and that will always be the same regardless of where the zero mark is. For ease of calculation we make it the ground level here.
12. (Original post by ghostwalker)
Yes.

You can assign any point as the zero height mark, since it's the difference in the GPE's that's important, and that will always be the same regardless of where the zero mark is. For ease of calculation we make it the ground level here.
Thankyou! Makes sense now - does the ‘work done’ change sign depending on direction or is it all positive?
13. (Original post by F3rnw3h)
Thankyou! Makes sense now - does the ‘work done’ change sign depending on direction or is it all positive?
If the "object" does work it's always positive.

If work is done to the object, e.g. being pulled by a force, then it's "energy input" or "work in" as notnek more correctly called it, and would be positive, BUT on the other side of the equation - see previous posts.

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