Hooke's Law: Solve Work Problem at MIT East Campus Dorms

  • Thread starter Thread starter shell4987
  • Start date Start date
  • Tags Tags
    Hooke's law Law
Click For Summary
SUMMARY

The discussion centers on a physics problem involving Hooke's Law, specifically applied to a catapult scenario at MIT's East Campus dorms. The spring constant of the surgical hose is given as 110 N/m, and the hose is stretched by 5.90 m. The correct calculation for the work done by the force from the hose on the balloon is derived from the elastic potential energy formula, resulting in a value of 1914.55 Joules. The negative sign in the potential energy calculation indicates the direction of the force, but the work done is considered as a positive value of 1914.55 Joules.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Knowledge of elastic potential energy calculations
  • Familiarity with the work-energy principle
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of elastic potential energy using the formula Ws = -1/2 kx²
  • Explore practical applications of Hooke's Law in engineering and physics
  • Learn about energy conservation in mechanical systems
  • Investigate the effects of varying spring constants on work done
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in practical applications of Hooke's Law in real-world scenarios.

shell4987
Messages
43
Reaction score
0

Homework Statement


During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law with a spring constant of 110 N/m. If the hose is stretched by 5.90 m and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length?


Homework Equations


F=-kx W=Fd


The Attempt at a Solution


I tried this problem and got -649 as the answer for the force, however I feel as if it is not correct, I also don't know what to do to get work, do I just multiply the -649 by 5.9 again to get work? Am I doing something wrong?
 
Physics news on Phys.org
Start by calculating the elastic potential energy stored in the streched pouch/hose.
 
Hootenanny said:
Start by calculating the elastic potential energy stored in the streched pouch/hose.

Okay, I did that and got 1914.55 as my answer, is that my answer? It seems as if I use Ws=-1/2kx^2 the negative (-1914.55) would be my answer and work? Am I right?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 14 ·
Replies
14
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
4
Views
2K