Question

A car moves at a constant speed on a straight but hilly road. One section has a crest and dip of the same $250m$ radius. As the car passes over the crest the normal force on the star is one half the $16kN$ weight of the car. What will be the force on the car as its passes through the bottom of the dip?

• A. $12kN$
• B. $24kN$
• C. $36kN$
• D. $48kN$

Answer 1

Answer: (B)

$24kN$

Here,

$mg-N_1=\frac{m{v}^2}{R}$

or $mg-mg/2=\frac{m{v}^2}{R}$

Therefore, $\frac{m{v}^2}{R}=\frac{mg}{2}=16kN/2=8kN$

Now, $N_2-mg=\frac{m{v}^2}{R}$

$N_2-mg=\frac{m{v}^2}{R}$

$N_2=mg+\frac{m{v}^2}{R}$

=$mg+\frac{mg}{2}+\frac{3mg}{2}$

=$\frac{3\left(16kN\right)}{2}$=24kN