A car of mass m = 1000 kg is moving with constant speed v = 10 m/s on a parabolic shaped bridge AFOE of span `l=40m` and height h = 20 m as shown in the figure. Then the net force applied by the bridge on the car when the car is at point F, is

Options

A. `5000 sqrt((5)/(2))N`
B. `(5000)/(sqrt(2))N`
C. `(10000)/(sqrt(2))N`
D. `5000 sqrt((2)/(5))N`

Answer

A

Explanation

Taking origin at O and OC as y-axis equation of bridge is `y=(x^(2))/(20)`

So, slope at `F=(dy)/(dx)=(2x)/(20)=(2xx10)/(20)=1`
At point F, `mg=(1)/(sqrt(2))-N=(mv^(2))/(R)`
`N=(mg)/(sqrt(2))-(mv^(2))/(R) and f_(s)=(mg)/(sqrt(2))`
At F, `R=((1+1)^(3//2))/((1)/(10)) = 20 sqrt(2)` (metre)
`N=(10m)/(sqrt(2))-(100m)/(20sqrt(2))=(5m)/(sqrt(2))`
`f_(s)=(20m)/(sqrt(2))`
`because ` Net force `=m sqrt((25)/(2)+(100)/(2))=5m sqrt((5)/(2))N`