Force and potential energy function

Click For Summary
SUMMARY

The discussion focuses on calculating the potential energy function U(x) associated with a conservative force defined as F = (-Ax + Bx^5) N, where A and B are constants. The correct potential energy function is derived by integrating the force with respect to x, leading to U(x) = (A/2)x^2 - (B/6)x^6 + C, with C determined by the condition U(0) = 0. Participants also explore the changes in potential and kinetic energy as the particle moves from x = 1.90 m to x = 3.80 m, ultimately finding that the change in potential energy is -977 J, indicating a corresponding increase in kinetic energy of 977 J.

PREREQUISITES
  • Understanding of conservative forces and potential energy concepts
  • Proficiency in calculus, specifically integration techniques
  • Familiarity with Newton's laws of motion
  • Knowledge of energy conservation principles
NEXT STEPS
  • Study integration techniques for deriving potential energy functions from force equations
  • Learn about conservative forces and their properties in classical mechanics
  • Explore energy conservation principles in mechanical systems
  • Review examples of potential energy calculations in various force fields
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone studying classical mechanics, particularly those focusing on energy concepts and force interactions.

bdh2991
Messages
102
Reaction score
0

Homework Statement



A single conservative force acting on a particle within a system varies as = (− Ax + Bx5) N, where A and B are constants, is in Newtons, and x is in meters.

(a) Calculate the potential energy function U(x) associated with this force, taking U = 0 at x = 0. (Use any variable or symbol stated above as necessary.)
U(x) =



(b) Find the change in potential energy and change in kinetic energy as the particle moves from x = 1.90 m to x = 3.80 m. (Use any variable or symbol stated above as necessary.)



Homework Equations



F = -∇U



The Attempt at a Solution



for part a i took the partial derivative with respect to x in order to get the potential energy function in which i got the answer: U(x) = (A - 5Bx^4)i the system said it was wrong though.


for part b i took my equation in part a to find the change in potential energy and got -977J
and i figured if the change in potential was lost 977 J then the change in kinetic energy would have to be gaining 977 Joules, but i was wrong again. please help
 
Physics news on Phys.org
Nvm i was doing it backwards and should have been integrating sorry guys
 
what was the answer

what was the answer to part A
 

Similar threads

Replies
15
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
9
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K