Answer:
[tex]v=10m/sec[/tex]
Explanation:
From the question we are told that
Radius of vertical r= 8m
Force exerted by passengers is 1/4 of weight
Generally the net force acting on top of the roller coaster is give to be
[tex]F_N+Fg[/tex]
where
[tex]F_N =forceof the normal[/tex]
[tex]Fg= force due to gravity[/tex]
Generally the net force is given to be [tex]FC(force towards center)[/tex]
[tex]F_C =F_N + Fg[/tex]
[tex]F_N =F_C -Fg[/tex]
[tex]F_N=F_C-F_g[/tex]
[tex]F_N=\frac{mv^2}{R} -mg[/tex]
Mathematical we can now derive V
[tex]m_g + \frac{8m}{4}= \frac{mv^2}{8}[/tex]
[tex]\frac{5mg}{4} =\frac{mv^2}{8}[/tex]
[tex]v^2 =\frac{40*10}{4}[/tex]
[tex]v=10m/sec[/tex]
Therefore the speed of the roller coaster is given ton be [tex]v=10m/sec[/tex]
[tex]v=10m/s[/tex]
Given:
Radius of vertical r= 8m
To find:
Force exerted by passengers is 1/4 of weight
Generally the net force acting on top of the roller coaster is give to be
[tex]F_N+F_g[/tex]
where
[tex]F_N=\text{Force of Normal}[/tex] and [tex]F_g=\text{Force due to gravity}[/tex]
Generally the net force is given to be [tex]FC- \text{Force towards centre}[/tex]
[tex]F_C=F_N+F_g\\\\F_N=F_C+F_g\\\\F_N=\frac{mv^2}{r} -mg[/tex]
Mathematically we can now derive V
[tex]mg-\frac{8m}{4} =\frac{mv^2}{8}\\\\\frac{5mg}{4}=\frac{mv^2}{8}\\\\v^2=\frac{40*10}{4} \\\\v=10m/s[/tex]
Therefore, the speed of the roller coaster is given to be 10m/s.
Learn more:
brainly.com/question/25834521
