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A piece of metal is released under water. The volume of the metal is 50.0 cm^3 and its specific gravity is 5.0. What is its initial acceleration?
I mostly just need some validation here because the answer isn't in the back of the book. I took the volume, converted to m^3. Using definition of s.g. solved for mass of object. Found the weight of object, which is equal to buoyant force. F=ma and solved for acceleration. Is that right?
 2 years ago
 2 years ago
A piece of metal is released under water. The volume of the metal is 50.0 cm^3 and its specific gravity is 5.0. What is its initial acceleration? I mostly just need some validation here because the answer isn't in the back of the book. I took the volume, converted to m^3. Using definition of s.g. solved for mass of object. Found the weight of object, which is equal to buoyant force. F=ma and solved for acceleration. Is that right?
 2 years ago
 2 years ago

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wcaprarBest ResponseYou've already chosen the best response.0
Hmmm, maybe not. I just end up with 9.8...
 2 years ago

fewscrewsmissingBest ResponseYou've already chosen the best response.1
The buoyant force a fluid applies to an object is equal to the weight of the volume of fluid displaced, not the weight of the object. I've just done this very quickly so it might not be entirely correct, but... Based on what we know of specific gravity and density and their respective equations, we can work out the mass of the object: \[\Large \begin{array}{l} SG = \frac{{{\rho _o}}}{{{\rho _f}}}\\ {\rho _f} = 1000{\rm{ kg/}}{{\rm{m}}^3}\\ {\rho _o} = SG \cdot {\rho _f} = 5.0 \cdot 1000 = 5000{\rm{ kg/}}{{\rm{m}}^3}\\ m = {\rho _o}V\\ = 5000 \cdot (50 \times {10^{  6}}) = 0.25{\rm{ kg}} \end{array}\] Note the conversion from g/cm^3 to kg/m^3 (10^6), and the assumption that the density of the water is 1000kg/m^3. We can work out the buoyancy force based on the density of water, gravity and the volume of displaced water: \[\Large \begin{array}{l} B = {\rho _f}gV\\ = 1000 \cdot 9.81 \cdot (50 \times {10^{  6}})\\ = 0.4905{\rm{ N}} \end{array}\] And the rest is simple forces: \[\Large \begin{array}{l} F = mg\\ = 0.25 \cdot 9.81 = {\rm{2}}{\rm{.4525 N}}\\ \Delta F = B  F\\ = 0.4905  2.4525\\ =  1.962{\rm{ N}}\\ F = ma\\ a = \frac{F}{m} = \frac{{  1.962}}{{0.25}} =  7.848{\rm{ m}}{{\rm{s}}^{2}} \end{array}\] Noting that the 'negative force' is simply relative to its direction, where negative would be down.
 2 years ago

wcaprarBest ResponseYou've already chosen the best response.0
Aha! The weight of the displaced water is where it got me! Thanks
 2 years ago
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