GPODT
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Regarding part a), why have they not included the work done against friction in the calculations? Surely some energy is 'lost' due to the work done against friction as the package slides down the rough plane?
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ztibor
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(Original post by GPODT)


Regarding part a), why have they not included the work done against friction in the calculations? Surely some energy is 'lost' due to the work done against friction as the package slides down the rough plane?
THe one of the main theorem of physics says that
changes in Energy = Work done
\Delta E= W
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GPODT
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(Original post by ztibor)
THe one of the main theorem of physics says that
changes in Energy = Work done
\Delta E= W
I know that the change in energy = Work done. But the energy lost due to friction is also part of this ''change in energy'' no?
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Killjoy-
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(Original post by GPODT)
I know that the change in energy = Work done. But the energy lost due to friction is also part of this ''change in energy'' no?
They have.
At S it has KE of 0.5m10^2 and GPE mgsin30
(If T is reference point of 0 GPE.)

Now at T KE is 0.5m9^2. Loss of GPE has not contributed to an increase in speed, rather more work than the value of GPE decrease has been done against friction.

So loss would be the difference in energies between points P and T. They have simply grouped the -9^2 into the KE equation when displaying the working.
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ztibor
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(Original post by GPODT)
I know that the change in energy = Work done. But the energy lost due to friction is also part of this ''change in energy'' no?
Yes it is part that change, increasing by acceleration and decreasing by friction, so the result of 2 works done is decreasing in velocity
\Delta E=\Delta E_m +\Delta E_p=W_{F_m}-W_{fric}+mg\Delta h=\frac{1}{2}m \Delta v^2+mg\Delta h
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