Resultant force of spring at the instant of release?

[Poptarts]

Homework Statement

What would the resultant force have been at the instant of release, if a 350 g mass were suspended from your spring, pulled down 3 cm and released?

What would have been the instantaneous acceleration at the moment of release?
The questions pertain to an experiment with a spring and slotted weights in a weight holder

Homework Equations

k=17080 dynes/cm
1) i. Resultant force F= -mg+kn
ii. Instantaneous acceleration a= f/m
or
2) i. Resultant force F= -k⋅s
ii. Instantaneous acceleration a= A x [1/√(M/k)]²

The Attempt at a Solution

I've found two ways to do both two problems but they each have different formulas and get different answers. Therefore, I don't know which one is right
1) i. F=-mg+kn {(-350g⋅980cm/s²)+(17080dynes/cm⋅3cm)}/(1x10⁵)= -2.9176N
ii. a=f/m=-2.9176N/0.350kg= -8.336m/s²
or
2) i. F= -k⋅s= (-17080dynes/cm⋅-3cm)/(1x10⁵)=.5124N
ii. a= A x [1/√(M/k)]²= 3cm ⋅{1/(√[350g+21.15g+(1/3)⋅3.30g]/17080dynes/cm)}²

I don't think 2) is the correct way but the unit conversions and two equations are kinda confusing me.
Also 2) contains more values from my experiment 21.15 g is the mass of the weight holder, 3.30g is the mass of the spring, which I don't think you need to solve this problem.

 

[TSny]

Welcome to PF!

The force that the spring exerts on the mass is kx where x is the amount the spring is stretched from its unstretched (natural) length. Was the spring already stretched some before pulling the mass down 3 cm?

 

[Poptarts]

Welcome to PF!

The force that the spring exerts on the mass is kx where x is the amount the spring is stretched from its unstretched (natural) length. Was the spring already stretched some before pulling the mass down 3 cm?

As I understand the question, no the spring was not stretched some before pulling the mass down 3 cm. However, the equation being k x makes more sense. The only experimental value that I think we were supposed to use is k, I guess the mass is there to confuse us.

 

[TSny]

The mass hangs at rest at the bottom of the spring before you pull it down 3 cm. While it hangs at rest, is the spring longer than its natural length?

 

[Poptarts]

The mass hangs at rest at the bottom of the spring before you pull it down 3 cm. While it hangs at rest, is the spring longer than its natural length?

I guess it would be? These are the exact typed questions, which aren't practiced during the experiment but just in the lab report. We never measured the affect mass has on spring vs. its natural length, so I don't think we need to take that into account. Also I figured out how to do the instantaneous acceleration and that's where the listed mass value is supposed to be used.
a=(-kx)/m (-17080dynes/cm⋅3cm)/(350g)= -146.4cm/s²

 

[TSny]

We never measured the affect mass has on spring vs. its natural length, so I don't think we need to take that into account.

I think that's a safe assumption. The mass of the spring is only about 1% of the hanging mass.

Also I figured out how to do the instantaneous acceleration and that's where the listed mass value is supposed to be used.
a=(-kx)/m (-17080dynes/cm⋅3cm)/(350g)= -146.4cm/s²

This looks correct except you should include the mass of the hanger as part of the accelerating mass.

If you want to see why your first approach isn't correct, look at what you wrote:

1) i. F=-mg+kn {(-350g⋅980cm/s²)+(17080dynes/cm⋅3cm)}/(1x10⁵)= -2.9176N
ii. a=f/m=-2.9176N/0.350kg= -8.336m/s²

The mass of the hanger needs to be included. Also, you do not have the correct number for the amount that the spring is stretched.

Can you figure out the distance x₀ in figure (B)? What is the total amount that the spring is stretched in (C)?