VI-1M (g)M2 (g)a(cm/s2)0 5.2 14.40 5.2 14.60 5.2 13.30 5.2 13.80 5.2 135 10.2 28.95 10.2 28.35 10.2 28.95 10.2 28.45 10.2 28.110 15.2 44.610 15.2 42.510 15.2 42.610 15.2 42.710 15.2 42.415 20.2 58.815 20.2 58.815 20.2 58.615 20.2 61.315 20.2 60.520 25.2 74.920 25.2 75.220 25.2 76.120 25.2 72.520 25.2 76.925 30.2 88.725 30.2 89.425 30.2 92.125 30.2 91.625 30.2 90.40 5 10 15 20 25 30 350102030405060708090100f(x) = 3.08x − 2.83 � vs 2 ��2 , g�, cm/s2From Excel, slope3.0830857-2.82728 interceptunc slope0.02534750.498132 unc interceptApplying Newton’s second law, we get a=(gM1+M2) M2The slope of the plot gives g as, s=gM1+M2 , where M1= (296.7+25) g g=s(M1+M2)=3.08308∗(296.7+25+5.2)=1007.85 cm/ s2Error analysis: σf ¿√(∂ f∂ xσx)2+…+… Substituting g for f and s for x,σg ¿√(∂ g∂ sσs)2=(∂(s∗(M1+M2))∂ sσs)=σs(M1+M2)Therefore, σg ¿0.0253475(296.7+25+5.2)cm/s2=8.286 cm/s2The range of g calculated lies between 1000 cm/s2 to 1016 cm/s2. The accepted value ofg= 980 cm/s2 is not in the range.g ±σg=100 8 ± 8 cm/s2VI-2The average of a when an impulse is given: a1=a1+a2+a3+a4+a5n=62.4+63.2+60.7+64.6+58.15=61.8 cm/s2And when the glider moves to the right,a2=58.8+58.8+58.6+61.3+60.55=59.6 cm/s2Applying Newton’s second law, we get f =(M1+M2) (a1−a2)2f =(296.7+25+5.2)g∗(61.8−59.6)cm/s22=359.59 g⋅c m/s2Smallest value for M2 is when M=0. Therefore,M2g=(5 .2 g)∗(1007.85 cm/s2)=5240.82 g ⋅ c m/s2The value of M2g is greater than the frictional force f.f = 360 g ⋅ c