(adsbygoogle = window.adsbygoogle || []).push({}); Max spped for a car to go around a curve track without skidding....confusing!!!

here's my question,

a car of mass 1000kg moves along the corner of a level road having a radius of curvature 35.0m. If the frictional force between the tyres and the road is 4kN, the maximum speed of the car without skidding at the corner is

A. 4.0 m/s

B. 8.8 m/s

C. 11.8 m/s

D. 140.0 m/s

ok, i knew there's a formula for the max spped of a car go around the curve track without skidding is v=(rag/h)^1/2, but is it possible to find the height form centre of gravity of car in the above case? I've tried usin' v=(Fr/m)^1/2 which is derived from the centripetal force, F=(mv^2)/r, nonetheless, if we using v=(Fr/m)^1/2, is this could be the maximum spped of car instead of using v=(rag/h)^1/2????

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# Max spped for a car to go around a curve track without skidding confusing

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