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Question Number 200428 by Spillover last updated on 18/Nov/23

Answered by ajfour last updated on 18/Nov/23

F=mg−((kv^2 )/2) ((vdv)/dx)=g−((kv^2 )/(2m))=−(k/(2m))(v^2 −((2mg)/k)) (1/2)∫_0 ^( v) ((2vdv)/(v^2 −((2mg)/k)))=−(k/(2m))∫_0 ^( x) dx ln [((((2mg)/k)−v^2 )/((((2mg)/k))))]=−((kx)/m) 1−((kv^2 )/(2mg))=e^(−((kx)/m)) v^2 =((2mg)/k)(1−e^(−((kx)/m)) ) (v^2 )∣_(max) =((2mg)/k) for x→∞

F=mg−kv22vdvdx=g−kv22m=−k2m(v2−2mgk)12∫0v2vdvv2−2mgk=−k2m∫0xdxln[2mgk−v2(2mgk)]=−kxm1−kv22mg=e−kxmv2=2mgk(1−e−kxm)(v2)∣max=2mgkforx→∞

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