A car moves at a constant speed on a straigh but hilly road. One section has a crest and dip of the `250m` radius.
(a) As the car passes over the crest the normal force on the car is half the `16kN` weight of the car. What will the noraml force on the car its passes through th ebottom of the dip?
(b) What is the greatest speed at which the car can move without leaving the road at the top of the hill ?
(c) Moving at a speed found in part (b) what will be the normal force on the car as it moves through the bottom of the dip? (Take, `g=10m//s^(2))`

Answer

A, B, C, D

Explanation

(a) `mg-N_(1)=(mv^(2))/(R)` ltbr4gt
or `mg-(mg)/(2)=(mv^(2))/(R)`
:. `(mv^(2))/(R)=(mg)/(2)=(16KN)/(2)=8KN`
Now, `N_(2)-mg=(mv^(2))/(R)` or `N_(2)=mg+(mv^(2))/(R)`
`=mg+(mg)/(2)+(3)/(2)mg`
`=(3)/(2)(16KN)=24KN`
(b) `mg-N=(mv^(2))/(R)` or `N=mg-(mv^(2))/(R)`
`0=mg-(mv_(max)^(2))/(R)`
:. `v_(max)=sqrt(gR)=sqrt(10xx250)=50m//s`
(c) `N_(2)-mg=(mv^(2))/(R)`
:. `N_(2)=m(g+(v^(2))/(R))`
`=(16xx10^(3))/(10)[10+(2500)/(250)]`
`=32xx10^(3)N=32KN`