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Question

In the given figure, m1=200 g, m2=600 g,μ1=0.3, μ2=0.8 and K=200 N/m.The minimum force that must be applied on m1 to just displace m2 is (in N). (Take g = 10 m/ s2)

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Solution


Ans: 3 N
Kx0=f12, Kx0=μ2m2g.
By work energy theorem for m1 and spring Fx0μ1m1gx012Kx20=0 F=μ1m1g+Kx02=(μ1m1+μ2m22)g=3N

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