Work and energy (change in potential)

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SUMMARY

The discussion focuses on the calculation of work and energy changes in a physics problem involving a constant force acting on a particle. The force vector is given as F = (3.02i + 5.45j) N, and the particle has a mass of 3.69 kg. The work done by the force when moving the particle to the position r = (1.97i - 2.77j) m is calculated as -9.15 J. The speed of the particle at this position is determined to be 2.71 m/s. The potential energy change, calculated using the work-energy principle, yields a result of 0.033 J, which raises questions about conservation of energy in a uniform force field.

PREREQUISITES
  • Understanding of vector operations in physics, specifically dot products.
  • Knowledge of the work-energy theorem and its application.
  • Familiarity with kinetic and potential energy calculations.
  • Basic concepts of conservative forces and energy conservation.
NEXT STEPS
  • Review the work-energy theorem and its implications for conservative forces.
  • Study the concept of potential energy in uniform force fields.
  • Practice solving similar problems involving work done by constant forces.
  • Explore the effects of rounding errors in energy calculations.
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of work and energy problems in a classroom setting.

lemin_rew
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Homework Statement



part a and b are solved(is it correct), but I am not sure how to solve part c.
A single constant force F = (3.02i + 5.45j)N acts on a 3.69kg particle.

a) Calculate the work done by this force if the particle moves from the origin to the point having the vector position r = (1.97i - 2.77j)m

b) What is the speed of the particle at r if its speed at the origin is 3.51m/s?

c)What is the change in the potential energy of the system?

Homework Equations




The Attempt at a Solution


a) F dot r = (3.02i + 5.45j) dot (1.97i +-2.77j) = (3.02x1.97)+(5.45x -2.77)= -9.15J

b)1/2mvi^2 - (F dot r) = 1/2mvf^2
(0.5x3.69kgx3.51^2 - 9.15J = 1/2(3.69kg)vf^2
v= 2.71m/s

c)W= Ek-Eg
-9.15J = 1/2mvf2-1/2mvi2 - Eg
-9.15J = 1/2(3.69kg)(2.71^2-3.51^2) - Eg
Eg= 0.033J
can anyone tell me what I am doing wrong for part c?
 
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I think the result you got in part (c) should have been zero. What you actually got looks like rounding error.
I would have thought that the lost KE has all gone into the PE (i.e. total conserved), no? A uniform force field would be conservative.
 

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