A 3.2 kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of this spring/mass sytem is 2.2 J. What is the elastic potential energy of the system when the 3.2 kg block is replaced by a 5.6 kg block?
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Physics - bobpursley, Monday, April 2, 2007 at 7:24am
PE= 1/2 kx^2 but W= kx or x= Weight/k
PE= 1/2 k (Weight/k)^2 reduce it.
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How do we find k. We have no x given.
Look above.
PE= 1/2 k x^2 and x= W/k
PE= 1/2 k W^2/k^2= 1/2 W^2/k
can you solve for k here? Now, knowing that, solve for x in the first equation.
Look at the last line in my original post: PE= 1/2 k (Weight/k)^2 reduce it.
You did not do that.
Now, there is another solution here. Notice that PE= 1/2 W^2/k. So PE is proportional to weight squared...so you can figure the new PE from the original problem statement (ratio of masses, original, and new).
You don't need to know what k is. For a fixed k, the energy is proportional to deflection^2 or weight^2. Therefore the full-deflection elastic potential energy increases by a factor of (5.6/3.2)^2 = 3.06