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(C) Given:Mass of the car,$m = 500 \, kg$Radius of the curved path,$r = 10 \, m$Speed of the car at point $P$,$v = 20 \, m/s$Acceleration due to gravi

Vedclass · Accepted Answer

$25$

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(C) Given:Mass of the car,$m = 500 \, kg$Radius of the curved path,$r = 10 \, m$Speed of the car at point $P$,$v = 20 \, m/s$Acceleration due to gravity,$g = 10 \, m/s^2$At the lowest point $P$,the forces acting on the car are the normal reaction $R$ (upwards) and the weight $mg$ (downwards).The net force providing the necessary centripetal force is $(R - mg)$.Therefore,$R - mg = \frac{mv^2}{r}$$R = mg + \frac{mv^2}{r}$Substituting the values:$R = (500 	imes 10) + \frac{500 	imes (20)^2}{10}$$R = 5000 + \frac{500 	imes 400}{10}$$R = 5000 + 20000$$R = 25000 \, N$Since $1 \, kN = 1000 \, N$,we have $R = 25 \, kN$.

$A$ car travelling on a smooth road passes through a curved portion of the road in the form of an arc of a circle of radius $10 \, m$. If the mass of the car is $500 \, kg$,the reaction on the car at the lowest point $P$ where its speed is $20 \, m/s$ is ......... $kN$.

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