Determing Spring Constant by Graphing

Click For Summary
SUMMARY

The discussion focuses on determining the spring constant (K) of a projectile launcher using experimental data from a lab involving a wooden cylinder and varying weights. The derived equation for K is K = 2Us/X02 = 2(.5m2ahfavg - mgh0) / X02. Participants are instructed to graph x02/m on the x-axis and hf-h0 on the y-axis, which should yield a linear relationship, with the slope and y-intercept providing critical insights into the spring constant. The relationship between potential energy and spring energy is emphasized, confirming the validity of the derived equation.

PREREQUISITES
  • Understanding of Hooke's Law and spring mechanics
  • Familiarity with energy conservation principles in physics
  • Basic knowledge of graphing linear relationships
  • Proficiency in using kinematic equations, specifically V2 = V02 + 2aΔy
NEXT STEPS
  • Learn how to graph linear equations in physics experiments
  • Study the derivation and application of Hooke's Law in practical scenarios
  • Explore energy conservation in mechanical systems
  • Investigate the relationship between mass, height, and potential energy in projectile motion
USEFUL FOR

Students in physics courses, educators teaching mechanics, and anyone involved in experimental physics or engineering projects related to spring dynamics and energy conservation.

dropgigawatt
Messages
10
Reaction score
0

Homework Statement


In class, we did a lab that consisted of putting a small wooden cylinder in a pipe that contained a spring, pushing it down on the spring and locking it in, then unlocking it which allowed it to launch straight up. We then would add fishing weights to the block to increase the mass and did 3 trials with each mass. This is my collected data:

Constant Values: h0 = .258 m, x0 = .022 m, xf = .000 m, V0 = 0 m/s, Vf = 0 m/s

Height of uncompressed projectile
Trials 1-3: m = .0228 kg, hf (avg) = 1.367 m
Trials 4-6: m = .0280 kg, hf (avg) = 1.304 m
Trials 7-9: m = .0329 kg, hf (avg) = 1.076 m
Trials 10-12: m = .0380 kg, hf (avg) = 0.951 m
Trials 13-15: m = .0424 kg, hf (avg) = 0.860 m

It now asks me to derive an equation for the spring constant of the projectile launcher based on the data I have. I came up with this: K = 2Us/X02 = 2(.5m2ahfavg - mgh0) / X02

Then it says Your goal for the lab is to determine the spring constant. If you graph your data with x02/m on the x-axis and hf-h0 on the y-axis you should get a straight line with a slope of _____ and a y-intercept of _____.

Homework Equations



V2 = V02 + 2aΔy
Us = .5Kx2

The Attempt at a Solution



As I said above, I derived K = 2Us/X02 = 2(.5m2ahfavg - mgh0) / X02

I tested it with some old data in my notes (that I knew K for) and it worked... but now that I see it wants me to graph x02/m on the x-axis and hf-h0 on the y-axis, I feel like I did it wrong because I can't see where those values became relevant.

So my question is did I derive that equation correctly? And how does it relate to what it wants me to graph? Thank you for your help!
 
Physics news on Phys.org
I would think that all the energy stored in the spring

Us = 1/2 K X[itex]^{2}_{o}[/itex]

would be converted to the potential energy of the block at the top

Ub = mg(hf - ho)

so that

hf - ho = [itex]( \frac{K}{2g} ) \frac{X^{2}_{o}}{m}[/itex]
 
Last edited:

Similar threads

  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
4K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K