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course PHY 202
1/19 4
Using water at about 70 Celsius we raised as much water as possible from the bottle to various heights. •Had we raised 20 milliliters of water to a height of 25 cm, by how much would the gravitational PE of the system have changed?
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PE=-mgy PE=1cm^3*1000cm/s^2*-25cm=-25000
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The work done against gravity is - m g `dy. The change in potential energy (which we denote `dPE) is equal and opposite to the work done against gravity, so dPE = m g `dy.
1 cm^3 is not the mass of 20 milliliters of water.
1000 cm/s^2 is a reasonable approximation of the 980 cm/s^2 acceleration of gravity, so that's OK.
The water is being raised so `dy = +25 cm, not -25 cm.
So basically you just have to adjust your signs, and use the mass of 20 milliliters of water, which is 20 grams.
You also need to express your result in the units that follow from the calculation. The units of your calculation as you presented it would be cm^3 * cm / s^2 * cm = cm^5 / s^2. However the cm^3 is not a unit of mass, so that will change.
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• Had we raised 15 milliliters of water to a height of 35 cm, by how much would the gravitational PE of the system have changed?
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PE=-mgy PE=1*1000*-35=35000
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• Had we raised 4 milliliters of water to a height of 80 cm, by how much would the gravitational PE of the system have changed?
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PE=-mgy PE=1*1000*-80=-80000
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• Based on these figures, to what approximate height do you estimate we would have had to raise water to maximize PE change?
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As high as you can!
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As you can see from the quantities given, the amount of water raised goes down as the height to which it is raised increases. This might well change your answer to this question.
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You've almost got these calculations, but you have to use the correct masses of the amounts of water raised. Different amounts of water have been raised, but you used the same quantity for the mass in every calculation.
You'll also want to adjust a couple of other details, as indicated in my note, but you're on the right track.
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