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Consider example 25, taking the coefficient of friction mu to be 0.4 and calculate the maximum compression of the spring (g=10ms^(-2)) |
Answer» Solution :Both the frictional force and the SPEING force act so as to oppose the compressin of the springas shown in figure.  Use the work-ENERGY theorem, The change in `KE=Delta K=K_(t)-K_(i)=0-1/2mv^(2)` The work done by the net force is `W=1/2kx_(m)^(2)MU mgx_(m)` Equating the two, we get `1/2mv^(2)=1/2Kx_(m)^(2)+mumgx_(m)` `mumg=0.4xx1500xx10=6000N` `Kx_(m)^(2)+2mumgx_(m)-mv^(2)=0""("After rearranging the given equlatin")` `x_(m)=(-2mumg+sqrt(4mu^(2)m^(2)g^(2)-4(k)(-mv^(2))))/(2K)("Taking + ve sign with square ROOT as"x_(m)"is +ve")` `=(-mumg+sqrt(mu^(2)m^(2)g^(2)+MKV^(2)))/(K)` `=(-0.4xx1500xx10+sqrt((0.4xx1500xx10)^(2)+1500+7.5xx10^(3)xx10^(2)))/(7.5xx10^(3))=3.74m` Whichas expected, is less than the result in Example 25. If the two forces on the body consist of a conservation for `F_(c)` and a non-nocervative force `F_(nc,)` the conservation of mechanical energy formula will have to be modified by the W-E theorem, `(F_(c)+F_(nc))Deltax=DeltaK` But `F_(c)Deltax=-DeltaV` Hence `Delta(K+V)=F_(nc)Deltax` `DeltaE=F_(nc) Delta x` Where E is the total mechanical energy. Over the path this assumes the form ` E_(f)-E_(i)=W_(nc)` Where `W_(nc)` is the total work done by the non-conservative forces over the path. Unlike conservative force, `W_(nc)` depends on the path taken. |
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