Question

A cyclist moves in a circular track of radius 50 metre.if coefficient of friction is 0.2 then find velcityof cyclist

Answer 1

Answer: velocity = √μ x r x g

                              =  √ 0.2 x 50 x 10

                              = √100

                               = 10 m/s

therefore the velocity of the cyclist is 10m/s

Explanation:

Answer 2

The velocity of cyclist is 9.89 m/s.

Explanation:

Given that,

Radius of the circular track, r = 50 m

The coefficient of friction, $\mu=0.2$

We need to find the velocity of cyclist. It is given by equating frictional force and the centripetal force, such that :

$v=\sqrt{\mu rg}$

$v=\sqrt{0.2\times 50\times 9.8}$

v = 9.89 m/s

So, the velocity of cyclist is 9.89 m/s. Hence, this is the required solution.

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