Question

If a cyclist moving with a speed of 4.9 m/s on a
level road can take a sharp circular turn of radius
4 m, then the coefficient of friction between the
cycle tyres and road is
(1) 0.51
(2) 0.41
(3) 0.71
(4) 0.61​

Answer 1

Explanation:

v = 4.9 m/s, r=4 m, g = 9.81

coefficient of friction = v^2/(r*g)

= (4.9^2)/(4*9.81) = 0.61

option 4 is correct

Answer 2

The coefficient of friction between the cycle tyres and the road is 0.61.

Explanation:

It is given that,

Speed of the cyclist, v = 4.9 m/s

Radius of the circular track, r = 4 m

On the banked curved, the coefficient of friction is given by balancing the centripetal force and the frictional force. So,

$\mu=\dfrac{v^2}{rg}$

$\mu=\dfrac{(4.9)^2}{4\times 9.8}$

$\mu=0.61$

So, the coefficient of friction between the cycle tyres and the road is 0.61. Hence, this is the required solution.

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