By how much is the energy stored in an ideal spring increased whcen it is stretch is increased from 8 to 16 cm

Question

By how much is the energy stored in an ideal spring increased when it is stretch is increased from 8 cm to 16cm?

Answer 1

Energy stored = k x^2, where x=stretch from the equilibrium position. Here k= spring constant. I think this was called Hooke's law for ideal spring.

If you mean before the spring was stretched by 8 cm from its normal (equilibrium) length, and now it's further stretched; then now the energy stored is (16/8)^2 = 4 times the former.

Answer 2

PE(delX = 8) = k/2 delX^2 = 64 k/2; PE(delX = 16) = 256 k/2. delX is the amount of stretch beyond the neutral point where PE(delX = 0) = 0. PE is the average force F/2 = k delX/2 acting over the stretch delX, which is why PE = k/2 delX^2.

Therefore PE(16) = PE(8) 256 k/2//64 k/2 = 4 PE(8); that is, you quadruple the potential energy when you double the stretch.

Note that the k's cancel out; so k is not a factor when comparing potential energies at different stretches or compressions of the same spring.

Answer 3

Hooke's Law: F = k x. Integrate this over the distance by which the spring is extended to yield E = (k x^2)/2

Answer 4

The problem can't be solved without to know its elastic constant k. The change in energy W(2)-W(1) is given by
W(2)-W(1)=k[x(1)^2-x(2)^2]/2. Look for k and solve the problem!

Answer 5

I would imagine that the increase would be by 100%.

If the spring is first stretched 0.08m, the the energy stored = (0.08 x 9.8m/s x mass)= 0.784 x mass in kg. Plug in 0.16m and everything gets doubled. So it increases by 100%.