Energy Conservation in a Vertical Spring System

[Matt1234]

Hello,
I need soe help please, i can't seem to get this one.

Homework Statement

A 0.20-kg mass is hung from a vertical spring of force constant 55 n/m. When the spring is released from its unstretched equilibrum position, the mass is allowed to fall. Use the law of conservation of energy to determine:
a) the speed of the mass after it falls 1.5cm
b) the distance the mass will fall before reversing direction

Books answers:
a) 0.48 m/s
b) 0.071m

Homework Equations

Eg = mgh
Ee =0.5 k x^2
Ek= 0.5 m v^2

The Attempt at a Solution

I have tried several things, but have erased them. I came up with 0.24 m/s for a) several times.

 

[cepheid]

The question asks you to use conservation of energy. You know that as the mass's height decreases, the system loses gravitational potential energy, (which is converted to kinetic energy). But by the same token, the system gains elastic potential energy because as the mass falls, the spring is elongated. Can you set up an equation for conservation of energy that takes both of these changes into account? It looks like you have all of the necessary equations.

For part b, the mass has to stop falling and come to rest before it changes direction and starts moving upwards. What is the reason why the mass would stop (come to rest)? The answer to that question should help you solve this part of the problem.

 

[Matt1234]

I believe i got part A

Eg = Ee + Ek

for part b

The mass would come to rest because the springs energy (in the up direction) is greater thatn the kinetic energy and gravitational potential.

I tried Ek = Etotal and that gave me the wrong answer so I am a tad confused.

 

[cepheid]

Rest means zero velocity. Which means zero kinetic energy. The mass will come to a stop because all of its kinetic energy has been converted into elastic potential energy. In your original equation:

Eg = Ee + Ek,

we have Ek = 0

 

[cepheid]

The mass would come to rest because the springs energy (in the up direction) is greater thatn the kinetic energy and gravitational potential.

Energy doesn't have a direction: it is a scalar, not a vector.

 

[Matt1234]

it would be negative though wouldn't it? Since the springs energy is opposing that of gravity, cost 180 = -1?

I ended up getting the answer i used the following:
Et = Eg + Ee

I left x as the unknowns and it ended up being quadratic, i then solved it for x to be 0.071 m, as the book says.

Im having a real hard time with this stuff, not sure why.

 

[cepheid]

It wasn't necessary to solve a quadratic. If you had done what I suggested in post #4 and set the kinetic energy to zero (because the mass has stopped moving), then you would have had:

mgx = ½kx²

2mg/k = x​

 

[Matt1234]

Thank you for your help.