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Question

A particle of mass 102 kg is moving along the positive x-axis under the influence of a force, F(x)=k2x2, where k=102Nm2. At time t=0, it is at x=1.0 m and its velocity v=0. The speed of the particle when it reaches x=05 m is

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Solution

Given, F(x)=k2x2m=102 kgk=102Nm2

a=Fm=k2x2×102=1022x2×102=12x2

a=dvdt=dvdx×dxdt=vdvdx=12x2

vdv=dx2x2

Integrating,

v0vdv=x1dx2x2

(at v=0,x=1)

v22=12[1x]x1=12[1x1]

At x=0.5,
v2=1v=±1 m/s

|v|=1 m/s

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