A car of 800 ,kg is taking a turn on a banked | vedclass

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(B) Given: Mass $m = 800 \,kg$, Radius $r = 300 \,m$, Angle of banking $	heta = 30^{\circ}$, Coefficient of static friction $\mu = 0.2$, Acceleration

Vedclass · Accepted Answer

$51.4$

Vedclass · Answer

(B) Given: Mass $m = 800 \,kg$, Radius $r = 300 \,m$, Angle of banking $	heta = 30^{\circ}$, Coefficient of static friction $\mu = 0.2$, Acceleration due to gravity $g = 10 \,m/s^2$.The formula for the maximum safe speed on a banked road with friction is given by:$V_{\max} = \sqrt{rg \left[ \frac{	an 	heta + \mu}{1 - \mu 	an 	heta} ight]}$Substitute the values:$V_{\max} = \sqrt{300 	imes 10 	imes \left[ \frac{	an 30^{\circ} + 0.2}{1 - 0.2 	imes 	an 30^{\circ}} ight]}$Since $	an 30^{\circ} = \frac{1}{\sqrt{3}} \approx \frac{1}{1.73} \approx 0.577$$V_{\max} = \sqrt{3000 	imes \left[ \frac{0.577 + 0.2}{1 - 0.2 	imes 0.577} ight]}$$V_{\max} = \sqrt{3000 	imes \left[ \frac{0.777}{0.8846} ight]}$$V_{\max} = \sqrt{3000 	imes 0.8783} \approx \sqrt{2635} \approx 51.33 \,m/s$Rounding to the nearest value, $V_{\max} \approx 51.4 \,m/s$.

$A$ car of $800 \,kg$ is taking a turn on a banked road of radius $300 \,m$ and angle of banking $30^{\circ}$. If the coefficient of static friction is $0.2$, then the maximum speed with which the car can negotiate the turn safely is: $(g=10 \,m/s^2, \sqrt{3}=1.73)$ (in $\,m/s$)

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