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P.E of a particle free to move to along xaxis is given by (x^3  x^2/2)J where x is in metre . The particle is initially kept at x=0 and then given a sight displacement x in +ve direction. The force acting on the particle will displace it to a Distance ?
 one year ago
 one year ago
Yahoo! Group Title
P.E of a particle free to move to along xaxis is given by (x^3  x^2/2)J where x is in metre . The particle is initially kept at x=0 and then given a sight displacement x in +ve direction. The force acting on the particle will displace it to a Distance ?
 one year ago
 one year ago

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Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
@experimentX @JFraser @siddhantsharan @gerryliyana
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
F =  d(PE)/dx
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
 (3x^2  x)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
F = 0 ... find the equilibrium
 one year ago
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