# MEI M2 Energy/Work Done

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#1
Is this the correct equation for work done?

WD = change in GPE + change in KE
WD = (final GPE - initial GPE) + (final KE - initial KE) (*)

Where WD is work done, GPE is gravitational potential energy and KE is kinetic energy.

I always seem to make errors with signs, so would applying the above equation (*) always result in the correct answer? WD would always be positive, wouldn't it?
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#2
Bump...
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5 years ago
#3
(Original post by lizard54142)
Is this the correct equation for work done?

WD = change in GPE + change in KE
WD = (final GPE - initial GPE) + (final KE - initial KE) (*)

Where WD is work done, GPE is gravitational potential energy and KE is kinetic energy.

I always seem to make errors with signs, so would applying the above equation (*) always result in the correct answer? WD would always be positive, wouldn't it?
Other way round in (*)

I tend to work with:

Energy before = energy after + work done

Assuming whatever it is, is doing the work, and work is not being done to it.

Then rearranging, we have WD = energy before - energy after.
Energy is going to be either GPE or KE at M2, I believe, and you sum over both of those.
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#4
(Original post by ghostwalker)
Other way round in (*)

I tend to work with:

Energy before = energy after + work done

Assuming whatever it is, is doing the work, and work is not being done to it.

Then rearranging, we have WD = energy before - energy after.
Energy is going to be either GPE or KE at M2, I believe, and you sum over both of those.

Would this equation work? I have to think in terms of changes in energy otherwise I won't get it imprinted in my mind. Also WD is the work done by external forces other than weight (i.e. resistance down a slope).

(Final KE - Initial KE) = (Final PE - Initial PE) - WD

EDIT: I think it should be (Final KE - Initial KE) = (Initial PE - Final PE) - WD instead?
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5 years ago
#5
(Original post by lizard54142)

Would this equation work? I have to think in terms of changes in energy otherwise I won't get it imprinted in my mind. Also WD is the work done by external forces other than weight (i.e. resistance down a slope).

(Final KE - Initial KE) = (Final PE - Initial PE) - WD

EDIT: I think it should be (Final KE - Initial KE) = (Initial PE - Final PE) - WD instead?
The first one won't be correct, since you have a final minus an initial, refering to the energy of the body, on each side of the equation.

The second one is a rearrangment of the equation I posted, breaking down "energy" into KE and PE.

However I was talking about work done by the body. If you want work done to the body then your original equation in your first post would cover it.
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#6
(Original post by ghostwalker)
The first one won't be correct, since you have a final minus an initial, refering to the energy of the body, on each side of the equation.

The second one is a rearrangment of the equation I posted, breaking down "energy" into KE and PE.

However I was talking about work done by the body. If you want work done to the body then your original equation in your first post would cover it.
I'll use a specific question example to help explain.

Part v) is asking you to calculate the final velocity, and hence see if it is greater than 3. I think that you need to use the equation in my "edit" for this.

So: 0.5x80x(v^2) - 0.5x80x(0.5^2) = (980-0) - ((392)/tan35) where v is unknown and to be calculated.
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5 years ago
#7
(Original post by lizard54142)
Part v) is asking you to calculate the final velocity, and hence see if it is greater than 3. I think that you need to use the equation in my "edit" for this.

So: 0.5x80x(v^2) - 0.5x80x(0.5^2) = (980-0) - ((392)/tan35) where v is unknown and to be calculated.
Looks reasonable.
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#8
(Original post by ghostwalker)
Looks reasonable.
Okay cool, so (Final KE - Initial KE) = (Initial PE - Final PE) - WD should work for the types of questions on this syllabus, I just have to be careful with the WD term. Cheers.
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