How Much Work to Tip a Cube into Unstable Equilibrium?

AI Thread Summary
To determine the minimum work needed to tip a 10.0-kg cube into unstable equilibrium, the focus is on calculating the change in potential energy as the cube shifts from its initial position to one where its center of gravity is directly above its edge. The correct distance for this calculation involves understanding the height change as the cube tips. The required work is found to be 5.08 J, which is derived from the energy needed to elevate the center of gravity during the tipping process. The key takeaway is that the problem emphasizes energy change rather than force. The discussion concludes with the successful calculation of the answer.
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A 10.0-kg solid uniform cube with 0.250-m sides rests on a level surface. What is the minimum amount of work necessary to put the cube in unstable equilibrium?

My understanding is that the cube has to be on one of its sides with it's center of gravity directly above it. However, I don't know how to find the force required to get the cube into this position. I'm not sure which is the correct distance the force has to travel through...i think it's either d = 0.250 * sin 45[degrees], or it's s = [pi]/4*0.250

The answer is 5.08 J, but I don't know how to arrive at this answer...
 
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The question is not about the force. You are being asked for the amount of energy required. All you have to do is calculate the change in potential energy from the initial orientation to the final orientation.
 
thanks! i got the answer
 

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