Question
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
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