We had worked this one out before but due to my bad note i really don't understand what is going on in the question. Following the question I will post what I wrote done. Please help me make some sense of it.
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 1.9 J. What is the elastic potential energy of the system when the 3.2 kg block is replaced by a 5.7 kg block?
PE for 3.2kg block is:
PE = 0.5(3.2)^2/ K
1.9J = 0.5(10.24)/K
1.9 = 5.12/K
K= 5.12/1.9
K= 2.695
PE for 5.7kg block is:
PE = 0.5(5.7)^2/ K
1.9J = 0.5(32.49)/K
1.9 = 16.245/K
K= 16.245/1.9
K= 8.55
therefore the elastic potential energy of the system due to the replacement is 8.55J.
K= 2.695
K stays the same. What changes is the deflection, X, which increases by a factor of 5.7/3.2 = 1.7813. The PE increases by the square of that factor, or 3.173. The new value of PE IS NOT 8.55 J. It is less than that.
there all rite
In the given question, we are asked to find the elastic potential energy of the system when the 3.2 kg block is replaced by a 5.7 kg block. Let's break down the steps you have written and make sense of them.
1. To find the elastic potential energy (PE) for the 3.2 kg block, you used the formula PE = 0.5 * (mass^2) / K, where K represents the spring constant.
- Substituting the values, you got PE = 0.5 * (3.2^2) / K.
- Simplifying further, you obtained 1.9 J = 0.5 * 10.24 / K.
- Solving for K, you divided both sides by 5.12 and got K = 2.695.
2. To find the elastic potential energy (PE) for the 5.7 kg block, you followed a similar process.
- Starting with the formula PE = 0.5 * (mass^2) / K, you replaced the mass with 5.7 kg.
- So, you got PE = 0.5 * (5.7^2) / K.
- Next, you equated this to the previously found value of 1.9 J and obtained 1.9 J = 0.5 * 32.49 / K.
- Solving for K, you divided both sides by 16.245 and found K = 8.55.
It seems that there is a misunderstanding in the last step. The statement "K stays the same" is incorrect. When you replace the 3.2 kg block with a 5.7 kg block, the spring constant will remain the same, but the deflection (or displacement) of the spring will change. This change in mass will affect the potential energy stored in the spring.
To determine the change in potential energy, you correctly identified that the change in mass is a factor of 5.7/3.2 = 1.7813. However, the potential energy changes by the square of that factor, not the linear factor. Therefore, the new potential energy for the system is not 8.55 J, but rather 1.9 J * (1.7813^2) = 6.708 J.
In summary, the correct elastic potential energy of the system when the 3.2 kg block is replaced by a 5.7 kg block is 6.708 J.