Deriving the work-energy theorem

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SUMMARY

The discussion centers on deriving the work-energy theorem, specifically for a particle of mass m moving with constant acceleration a. The user attempts to express acceleration in terms of initial velocity (v_initial), final velocity (v_final), and displacement (s) using the formula (v_final - v_initial)^2/s. However, the response highlights that this formula is flawed as it does not account for negative acceleration and emphasizes the importance of considering units and the definition of acceleration as the change in velocity per unit time.

PREREQUISITES
  • Understanding of basic mechanics concepts, including velocity and acceleration.
  • Familiarity with kinematic equations for motion in one dimension.
  • Knowledge of the work-energy theorem and its implications in physics.
  • Ability to manipulate algebraic expressions and units in physics problems.
NEXT STEPS
  • Review kinematic equations, particularly those relating velocity, acceleration, and displacement.
  • Study the derivation and applications of the work-energy theorem in mechanics.
  • Explore examples of motion with both positive and negative acceleration.
  • Practice solving problems involving the calculation of acceleration from given variables.
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of motion and energy relationships in classical mechanics.

badman
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In this problem, you will use your prior knowledge to derive one of the most important relationships in mechanics: the work-energy theorem. We will start with a special case: a particle of mass m moving in the x direction at constant acceleration a. During a certain interval of time, the particle accelerates from v_initial to v_final, undergoing displacement s given by s=x_{\rm final}-x_{\rm initial}.


the problems asks me to find the acceleration :Find the acceleration a of the particle.
Express the acceleration in terms of v_initial, v_final, and s.
this is my answer: (v_final-v_initial)^2/s

but it says that: Your answer is close. One problem with your answer is that it is always positive, but acceleration can also be negative. ?
 
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Look at the units, and remember acceleration is the change in velocity per unit time.
 
Your answer is incorrect. Show how you arrived at it.
 

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