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
To find the elastic potential energy of the system when the 3.2 kg block is replaced by a 5.6 kg block, we can use the fact that the elastic potential energy is proportional to the weight squared.
First, let's find the proportionality constant, k. We have no value for x (deflection), but we can use the relation W = kx, where W is the weight of the block. Rearranging for x, we have x = W/k.
Now, let's substitute this value of x into the equation for elastic potential energy: PE = (1/2)kx^2.
PE = (1/2)k(W/k)^2, which simplifies to PE = (1/2)W^2/k.
From this equation, we can see that PE is proportional to the weight squared. Therefore, the change in PE is determined by the ratio of weights of the two blocks.
To find the new PE, we can calculate the ratio of masses, 5.6 kg (new block) to 3.2 kg (original block), which is (5.6/3.2)^2.
This ratio gives us the factor by which the original PE needs to be multiplied to get the new PE.
Therefore, the elastic potential energy of the system when the 3.2 kg block is replaced by a 5.6 kg block is 3.06 times the original PE.