The potential energy in the Spring?

Question

A 1.2kg object hanging from a spring of force constant 305N/m oscillates with a maximum speed of 33cm/s. (Choose U=0 when the object is in equilbrium). If the maximum displacement is 2.1cm, the total energy of the system is 0.065J and the gravitational potential energy is -0.247J.

What is the potential energy in the spring?

The potential energy in the spring should be greater than the total energy of the system, but the answer should be around 0.372J. I don't know how to get the exact value. So if anyone knows, please answer.

Answer 1

There's a pretty simple way to solve this question if you make a few simplifying assumptions:

1. The spring itself is approximately massless
2. Air resistance and other forms of damping are negligible
3. The equilibrium point of the system is the point at which the mass hangs on the spring when it is not oscillating.

Then you can use the definitions of gravitational potential and spring potential to solve for the latter:

U = m * g * (x - xeq)

S = (1/2) * k * (x - xeq)²

where U is gravitational potential, S is spring potential, m is the block's mass, g is the gravitational acceleration, k is the spring constant and (x - xeq) is the block's distance from the equilibrium position.

We can find the distance (x-xeq) by using the gravitational potential equation:

(x - xeq) = U / (mg)

We then substitute this expression into the spring potential equation:

S = (1/2) * k * (U / (mg))²

Finally, we substitute the known values to find:

S = (1/2) * 305 * (-.247 / (1.2 * 9.8))²

S = .06727 J

Thus, there is .06727 J of potential energy stored in the spring. NOTE: The total energy in the system is comprised of the kinetic and potential energies of the mass, and the potential stored in the spring. The fact that there is more potential energy in the spring than in the total energy of the system is because there is still .2447 J of kinetic energy in the mass which is unaccounted for in this problem. If you sum up the gravitational potential, spring potential and this kinetic energy, you will get the system's total mass.

Answer 2

The potential energy of the spring when it is at rest with the weight hung but not oscillating is (1/2)k(ext1)^2. ext1 = (1.2x9.8)/305 m = 0.0386 m. Therefore, the total extension, ext2 of the spring when the hung object is at its extreme position is given by
ext2 = 0.0386+0.021 = 0.0596. So the change in the mechanical potential energy, U of the spring is
U = (1/2)k[(ext2)^2 - (ext1)^2] = 0.5x305(5.96^2 - 3.86^2) = 0.313
Thus the total energy, TE when the mass is at its lowest position becomes:

TE = 0.313 - 0.247 = 0.066, which matches with the total energy of the system given in the problem.

The formula 1/2kx^2 applies only when x is the extension of the spring measured from its tension free state.

Answer 3

U=0.5 * k x^2

do the math