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Force acts on a particle

  1. Oct 15, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle moving in the xy plane undergoes a displacement ⃗s = (sx ˆi + sy ˆj), with sx = 1.74 m, sy = 4.3 m, while a constant force
    F⃗ = (Fxˆi+Fyˆj), with Fx = 5.1 N, Fy = 1.95 N, acts on the particle.
    Calculate the magnitude of the displacement.
    Answer in units of m.

    Find the magnitude of the force. Answer in units of N.

    Calculate the work done by F⃗ . Answer in units of J.

    Calculate the angle between F⃗ and ⃗s. Answer in units of degrees.

    2. Relevant equations

    3. The attempt at a solution

    i took the dot product of the two thinking i would get the magnitude of the displacement. im lost as to why it is incorrect?

    1.74*5.1 + 4.3*1.95 = 17.259
     
  2. jcsd
  3. Oct 15, 2010 #2
    its simple, i figured it out... i feel like an idiot

    pythagorean theorm, magnitude = sqrt(1.74^2 + 4.3^2)
     
  4. Oct 15, 2010 #3
    the entire problem is simple, i figured it out so i thought i would post it in case anyone would stumble upon this in the future

    Find the magnitude of the force. Answer in units of N.

    this is just the length force vector, duh...

    Calculate the work done by F⃗ . Answer in units of J.

    this is the dot product of the two

    Calculate the angle between F⃗ and ⃗s. Answer in units of degrees.

    this angle is found by taking the inverse of any trig function, provided that you use the correct sides that coorespond to that function, of the larger angle, - the smaller angle.
     
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