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DanDanDan
A block with mass of 0.5 kg is forced against a horizontal spring of negligible mass. compressing the spring a distance of 0.20m. When released, the block moves on a horizontal tabletop for 1 meter before coming to rest. The spring constant is 100N/m. What is the coefficient of friction between the block and the tabletop??
first calculate the potential energy stored in the spring. which is \[1/2kx^{2}\] . when the block is in rest it has zero mechanical energy,this means friction has done work equal to that of initial potential energy. therefore, F(friction) x X(distance moved) = \[1/2kx^{2}\] uN x 1 = 2 u = 2/5
friction foced on block,lonely, when it separeted from spring :\[-\mu _{s}mg=ma\] i think we use conservation energy between two points: where spring is completly compressed & when block release from it,so yields: \[(1/2)kx ^{2}=(1/2)mv ^{2}\] from above obtained v^2. (this (v) change to v0 for the next part of motion) independent equation of motion yeild's a : \[v ^{2}-v _{0}^{2}=2ax\] at rest v=0 so \[a=-v _{0}^{2}/2x\](x=1,according to ask) now if insert "a" in newton 2nd law \[\mu _{s}\] equals to 0.4
sorry \[\mu _{k}\] is correct, not \[\mu _{s}\]