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Q.

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 16kN 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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a

54kN

b

24kN

c

45kN

d

39.6rpm

answer is C.

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Detailed Solution

mg-N1=mv2R

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 or mg-mg2=mv2R

∴  mv2R =mg2=16kN2=8kN

 Now, N2-mg =mv2R or N2=mg+mv2R

 =mg+mg2=32mg

=32(16kN)=24kN

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