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anastassius [24]
3 years ago
14

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
Physics
1 answer:
lorasvet [3.4K]3 years ago
4 0

Answer:

F_N>F_N'

Explanation:

From the question we are told that

Radius of curvaturer=100m

Mass M=300kg

initial Speed of Motorcycle  V_1=30m/s

Final  Speed of Motorcycle  V_2=33m/s

Generally the equation Force at initial speed is  mathematically given as

F_N=mg-\frac{mv^2}{R}

F_N=300*9.8-\frac{(300*30)^2}{100}

F_N=240N

Generally the equation Force at Final speed is  mathematically given as

F_N'=mg-\frac{mv'^2}{R}

F_N'=300*9.8-\frac{(300*33)^2}{100}

F_N'=-327N

Therefore

F_N>F_N'

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