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Answer:
[tex]F_N>F_N'[/tex]
Explanation:
From the question we are told that
Radius of curvature[tex]r=100m[/tex]
Mass [tex]M=300kg[/tex]
initial Speed of Motorcycle [tex]V_1=30m/s[/tex]
Final Speed of Motorcycle [tex]V_2=33m/s[/tex]
Generally the equation Force at initial speed is mathematically given as
[tex]F_N=mg-\frac{mv^2}{R}[/tex]
[tex]F_N=300*9.8-\frac{(300*30)^2}{100}[/tex]
[tex]F_N=240N[/tex]
Generally the equation Force at Final speed is mathematically given as
[tex]F_N'=mg-\frac{mv'^2}{R}[/tex]
[tex]F_N'=300*9.8-\frac{(300*33)^2}{100}[/tex]
[tex]F_N'=-327N[/tex]
Therefore
[tex]F_N>F_N'[/tex]
Following are the calculation to the given question:
Solution:
Using formula:
[tex]\to mg - F^{'}_N=\frac{mv^{2}}{R} \\[/tex]
Calculating the Initial value:
[tex]\to F_N = mg - \frac{mv^2}{R}\\\\[/tex]
[tex]= 300 \times (9.8 - \frac{(30)^2}{100}) \\\\= 300 \times (9.8 - \frac{900}{100}) \\\\= 300 \times (9.8 - 9) \\\\= 300 \times (0.8) \\\\ = 240\ N \\\\[/tex]
Calculating the Final value:
[tex]\to F^{'}_{N}= 300 (9.8 -\frac{33^2}{100})\\\\[/tex]
[tex]= 300 (9.8 -\frac{1089}{100})\\\\= 300 (9.8 - 10.89)\\\\= 300 (- 1.09)\\\\=-327[/tex]
Therefore, the answer is "the new normal force is less than [tex]F_{N}[/tex]" or [tex]\bold{F^{'}_{N}< F_{N}}[/tex].
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