JEE PYQ: Laws Of Motion Question 6
Question 6 - 2021 | 17 Mar Shift 1] (MCQ)
A modern grand-prix racing car of mass m is travelling on a flat track in a circular arc of radius R with a speed v. If the coefficient of static friction between the tyres and the track is $\mu_s$, then the magnitude of negative lift $F_L$ acting downwards on the car is :
(1) $m\left(\frac{v^2}{\mu_s R} + g\right)$
(2) $m\left(\frac{v^2}{\mu_s R} - g\right)$ [Correct]
(3) $m\left(g - \frac{v^2}{\mu_s R}\right)$
(4) $-m\left(g + \frac{v^2}{\mu_s R}\right)$
Show Answer
Answer: (2)
Solution
$\mu_s N = \frac{mv^2}{R}$, $N = mg + F_L$. So $F_L = \frac{mv^2}{\mu_s R} - mg = m\left(\frac{v^2}{\mu_s R} - g\right)$.
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