Solving a Simple Spring Problem: Finding the Speed of a Launched Object

  • Thread starter Thread starter Rha1828
  • Start date Start date
  • Tags Tags
    Spring
Click For Summary
SUMMARY

The discussion centers on calculating the speed of an object launched from a spring gun, with a mass of 2.1 kg and a spring constant of 19 N/m, using a compression of 3.6 cm (0.036 m). The correct approach involves equating the potential energy stored in the spring to the kinetic energy of the object. The formula used is ½kx² = ½mv², leading to the conclusion that the speed can be derived by rearranging this equation to solve for v.

PREREQUISITES
  • Understanding of potential energy and kinetic energy concepts
  • Familiarity with Hooke's Law and spring constants
  • Basic algebra for rearranging equations
  • Knowledge of unit conversions (e.g., cm to m)
NEXT STEPS
  • Study the derivation of the kinetic energy formula (KE = ½mv²)
  • Learn about Hooke's Law and its applications in physics
  • Explore energy conservation principles in mechanical systems
  • Practice solving problems involving springs and projectile motion
USEFUL FOR

Students in physics, educators teaching mechanics, and anyone interested in understanding energy transformations in spring systems.

Rha1828
Messages
11
Reaction score
0
Simple Spring Problem Help!

A spring gun fires an object of mass 2.1 kg horizontally across a frictionless table. The spring has a spring constant of 19 N/m and is initially compressed by 3.6 cm. What is the speed of the object as it leaves the spring gun?

I can not figure out this problem.
3.6cm = 0.036 m
then I tired (0.036)(19) then dividing that by the mass
nothing seems to be working

Any help or direction would be great! Thank you!
 
Physics news on Phys.org


Typically you would want to consider the potential energy of the compressed spring being converted to kinetic energy.
 


It sounds like you are just doing things at random to see if they produce the answer in the back of the book. It would be far better to set the problem up in symbols and work it out in what you think is a rational manner before you ever substitute any numbers at all. You really need to be thinking about what you are doing and why.
 

Similar threads

Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
22
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
10K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K