course Phy 232
7/7 2:30 pm
• How much potential energy gain was therefore accomplished, per cm^3 of water raised?Potential energy=mass*gravity*height
I’m not quite sure what this question is asking.
Gravitational pull= 9.8 m/s^2
Height= 35 cm=0.35 m
denisty of water=1.0 g/cm^3
d=m/v
m=d*v=1.0 g/cm^3*1 cm^3=1.0 g
PE=1.0 g *9.8 m/s^2*0.35 m= 3.43 J
the change in potential energy is the work done against gravity
you know the density, and hence the mass of 1 cm^3 of water
so you can find the force exerted by gravity on the water, which you can use to calculate the work done against gravity
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• A cupful of water has a volume of about 250 cm^3 (very roughly). By how much would the potential energy of the system therefore be increased if you raised a cupful of water to the height of the outflow position?
The potential energy would increase by a factor of the difference of the height of the water to the outflow position because Potential Energy=m*g*h.
The mass can be calculated by using the volume and density (mass=density*volume). The gravitational pull is 9.8 m/s^2 and h is the height of the outflow position.
d=m/v
m=d*v=1.0 g/cm^3*250 cm^3= 250 g
PE=250 g*9.8 m/s^2*h
PE=2450* h
are we assuming the height is again 35 cm?
PE=857.5 J
Right. You can give a specific quantitative answer to this question.
Good, but grams * m/s^2 * m = gram m^2 / s^2, which is not Joules.
A Joule is a Newton * meter = kg m/s^2 * m = kg m^2 / s^2.
If you express your 250 grams as .25 kg, you will get the result in Joules.
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