Law of Conservation of Energy- mass and spring

[Maiia]

Homework Statement

A 2300g mass starts from rest and slides a distance L down a frictionless 26 degree incline, where it contacts an unstressed 34cm long spring of negligible mass as shown in the figure. The mass slides an additional 17cm as it is brought momentarily to rest by compressing the spring of force constant 19N/cm. The acceleration of gravity is 9.8m/s^2. Note: The spring lies along the surface of the ramp. Assume the ramp is frictionless. Now, the external force is rapidly removed so that the compressed spring can push up the mass. Find the initial separation L between mass and spring. Answer in units of m.

Picture:

I was wondering if someone could check to see if the way I am approaching this problem is correct.

There are three energies: Gravitational PE, Elastic PE and Kinetic Energy. When you set them equal to each other:
(1/2)mv^2 + mgy + (1/2)kx^2= (1/2)mv2^2 + mgy2+ (1/2)kx^2
if the left side of the equation is before the object slides down the ramp and the right side of the equation is after the spring is compressed, then it simplifies to:
mgy= mgy2 + (1/2)kx^2 because there is no KE or EPE at the beginning and at the end, there is no KE (gets converted to PE and EPE)
So, inputting -Y for the amount the spring is compressed and h for the height of the ramp:
mgh= -mgY + (1/2)kY^2
(-mgY+ (1/2)kY^2)/mg
Plugging in the numbers:
(-2.3 * 9.8 * .17m + .5 * 1900 N/m * .17^2)/ 2.3 * 9.8
I get h to equal 1.048056708m
Going back to the equation Work Done= Fcosthetad which is equal to mgy, then shouldn't the distance d be equal to h/costheta?

 

[tiny-tim]

Hi Maiia!

A 2300g mass starts from rest and slides a distance L down a frictionless 26 degree incline, where it contacts an unstressed 34cm long spring of negligible mass as shown in the figure. The mass slides an additional 17cm as it is brought momentarily to rest by compressing the spring of force constant 19N/cm. The acceleration of gravity is 9.8m/s^2. Note: The spring lies along the surface of the ramp. Assume the ramp is frictionless. Now, the external force is rapidly removed so that the compressed spring can push up the mass. Find the initial separation L between mass and spring. Answer in units of m.
…
mgh= -mgY + (1/2)kY^2
(-mgY+ (1/2)kY^2)/mg
Plugging in the numbers:
(-2.3 * 9.8 * .17m + .5 * 1900 N/m * .17^2)/ 2.3 * 9.8

You were doing fine until just before these lines …

i] it isn't mgY, because you need the vertical distance

ii] this is a quadratic equation …

 

[Maiia]

oh i see.. thanks! :)