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

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$54 kN$

$24 kN$

$45 kN$

$39.6 rpm$

answer is C.

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Detailed Solution

$m g - N_{1} = \frac{m v^{2}}{R}$

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

$∴ \frac{m v^{2}}{R} = \frac{m g}{2} = \frac{16 kN}{2} = 8 kN$

$Now, N_{2} - m g = \frac{m v^{2}}{R} or N_{2} = m g + \frac{m v^{2}}{R}$

$= m g + \frac{m g}{2} = \frac{3}{2} m g$

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

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