The potential energy of a particle of mass 5 Kg moving in x-y plane is given as U = (7x + 24y ) j , x and y being in metre. Initially at t=0 , the particle is at the origin ( 0, 0 ) moving  with a velocity of ( 8.6i + 23.2j ) ms-1 . Then 
  1. the velocity of the particle at t = 4s , is 5 ms-1
  2. the acceleration of the particle is 5 ms-2
  3. the direction of motion of the particle initially is at right angles to the direction of acceleration
  4. the path of the particle is circle 

Dear Student,

potential energy, U=7x+24y Jmass, m=5 kgFx=-dUdxFx=-7xFy=-dUdyFy=-24F=Fx2+Fy2F=-72+-242F=49+576F=625=25 Nacceleration, a=Fma=255=5 m/s2

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