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A particle of mass 10² Kg is moving along the positive x-axis under the influence of a force F(x) = -k/2x² where k= 10-² NM² . at time t=0 it is at x=1.0 m and its velocity is v=0 (a) Find its velocity when it reaches x= 0.5 m

GUNAJIT KUMAR BHUYAN , 10 Years ago
Grade 11
anser 1 Answers
Shivam

Last Activity: 10 Years ago

Change in Kinetic energy = work done 
½ mv2 – 1/2 mu2 =∫ F.dx =>1/2 m(vf2-v2) = [k/x3]]
½ * 102 (vf2 )= 10-2/(1/.125 – 1/1)
vf2= 14 * 10-4
vf = 3.74 * 10-2 m/s
 
                             

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