How High Does the Dart Travel When the Spring is Compressed Half as Far?

  • Thread starter Thread starter preluderacer
  • Start date Start date
  • Tags Tags
    Gun Spring
Click For Summary
SUMMARY

The discussion focuses on calculating the height a dart reaches when shot from a spring-loaded dart gun, specifically when the spring is compressed half as far compared to a previous shot that achieved a height of 24 meters. Using the formula for elastic potential energy (EPE = 0.5kx²) and gravitational potential energy (GPE = mgh), it is established that compressing the spring to half its original distance results in a reduction of potential energy to one-fourth, leading to a maximum height of 6 meters for the second shot. This conclusion is derived from the direct proportionality between the potential energy stored in the spring and the height achieved by the dart.

PREREQUISITES
  • Understanding of Hooke's Law and spring mechanics
  • Knowledge of gravitational potential energy calculations
  • Familiarity with energy conservation principles
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the relationship between elastic potential energy and gravitational potential energy
  • Learn about the implications of Hooke's Law in real-world applications
  • Explore more complex spring systems and their energy transformations
  • Investigate the effects of friction and air resistance on projectile motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for practical examples of spring dynamics and energy transfer concepts.

preluderacer
Messages
83
Reaction score
0

Homework Statement


A spring-loaded dart gun is used to shoot a dart straight up into the air, and the dart reaches a maximum height of 24 meters. The same dart is shot up a second time from the same gun, but this time the spring is compressed only half as far (compared to the first shot). How far up does the dart go this time (neglect friction and assume the spring obeys Hooke's law)?


The Attempt at a Solution



I know I have to use 1/2(k)x^2 , but I am not quite sure how to set this problem up.
 
Physics news on Phys.org
All of your elastic potential energy is going to be transferred to gravitational potential energy; GPE = mgh. Your elastic potential energy, like you mentioned, is going to be 0.5kx^2. Look at the ratios that form; if the spring is compressed only half way, the potential energy is 1/4 of what it was when pulled back fully - which would proportionally change the height.

That makes sense in my head...it's hard to form into words, so I hope that somewhat makes sense!
 
So it would only travel 6 meters?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
Replies
17
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
7K
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
Replies
17
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K