- #1
Abu
Homework Statement
A 1.00kg mass and 2.00kg mass are set gently on a platform mounted on an ideal spring of force constant 40.0 N/m. The 2.00 kg mass is suddenly removed. How high above its starting position does the 1.00 kg mass reach?
Related to it... An 87 g box is attached to a spring with a force constant of 82 N/m. The spring is compressed 11cm and the system is released.
A) What is the speed of the box when the spring is stretched by 7.0 cm?
B) What is the maximum speed of the box?
Homework Equations
Elastic potential energy = 1/2kx^2
Kinetic energy = 1/2mv^2
Potential energy = mgh
The Attempt at a Solution
The first problem is what confused me the most. It confused me because I don't know whether it is a case of simple harmonic motion or not. If you remove the 2 kg mass, it will move upwards, but will it come back down and oscillate or remain at the top at a new equilibrium? I will show you what I did:
mg = -kx
3(9.8) = -40x
x = -0.735
Now here is my problem, I know that you use x for the elastic potential energy 1/2kx^2, but what is the elastic potential energy equal to? Once you remove the 2kg mass, the other mass will move up. If it reaches equilibrium at the top right away, then the elastic potential energy at the bottom should transfer to just potential energy mgh, since it is not moving anymore and it is at a height? But if the mass does not stop at the top and instead comes back down, shouldn't the elastic potential energy at the bottom equal the elastic potential energy at the top?
So its either:
1/2kx^2 = mgh
OR
1/2kx^2 = 1/2kx^2 where the first x is 0.735?
I think I am confusing myself.. can someone explain this problem to me?
And for the second problem, all I need to know is if the 11 cm the amplitude? I mean the maximum displacement in the simple harmonic motion.
I know I am asking for a lot, thank you all anyways.