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A motorcycle passes over the top of a hill that has a radius of curvature of 100 m. The mass of the motorcycle plus rider is 300 kg. The motorcycle is moving at a speed of 30 m/s. The surface exerts a normal force of magnitude FN on the motorcycle.The motorcycle passes over the top of the hill again but now is moving at a speed of 33 m/s. How does the new normal force exerted on the motorcycle compare to FN

Sagot :

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].

Learn more:

brainly.com/question/12910909

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