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Force and Potential energy funtion

  1. Mar 21, 2008 #1
    1. The so-called Yukawa potential
    (attached)
    gives a fairly accurate description of the interaction between nucleons[that is, neutrons and protons, the constituents of the nucleus]. The constants r0 = 1.5x10-15metres and U0 = 50 MeV.[a] Find the corresponding expression for the force of attraction.


    2. Fr= -dU/dr



    3. When i finnaly calculate it, the answer is supposed to be negative because
    Fr= - dU/dr
    = - d (U is given negative)/dr
    { -ve and -ve cancels out }

    Fr= +ve d(U)/dr
    after integrating U, a -ve sign appears...
    so...shouldnt the answer be negative? but the actual solution is a positive. can u tell me y ?
     

    Attached Files:

  2. jcsd
  3. Mar 21, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    When you find the force by via -dU/dr, the resulting sign will tell you which way the force points (with respect to the variable r). Since it's an attractive force, that will be negative. The magnitude of the force of attraction will be positive, of course.
     
  4. Mar 21, 2008 #3
    thanks for the reply
     
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