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Grade: 11

                        

A particle of mass 2 kg is subjected to a two dimensional Conservative force given by F x = (-2x+2y) , F y = (2x-y²) (x, y in m and F in N). If particle has kinetic energy of 8/3 J at point (2,3), find speed (in m/s) of the particle when it reaches (1,2)

2 years ago

Answers : (1)

Raj150
13 Points
							
given F=(-2x+2y)i^+(2x-y2)j^
         KEi=8/3J
let S=(dx)i^+(dy)j^ 
   Work Done=Change in K.E
              F.S=(1/2)m(v2-u2)
  •           ∫[(-2x+2y)i^+(2x-y2)j^][(dx)i^+(dy)j^]=v2-64/9
  •          ∫(-2x+2y)dx+(2x-y2)dy=v2-64/9
  •         21-2xdx+32-y2dy+∫2ydx+∫2xdy=v2-64/9
  •         3+19/8+(2,3)(1,2)d(2xy)=v2-64/9
  •         28/3+4-12=v2-64/9
        After Solving The Above Equation We Get
                              V=2.905m/s
      
7 months ago
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