contestada

what is the maximum speed with which a 1200- kg car can round a turn of radius 92.0 m on a flat road if the coefficient of static friction between tires and road is 0.40?

Respuesta :

The maximum speed with which a 1200- kg car can round a turn of radius 92.0 m on a flat road if the coefficient of static friction between tires and road is 0.40 will be 18.6 m/s.

A force that holds an object at rest is called static friction. The definition of static friction is: The resistance people feel when they attempt to move a stationary object across a surface without actually causing any relative motion between their body and the surface they are moving the object across.

The frictional force acting on the car is given by:

[tex]F_{f} =umg =(0.40) (1200kg)(9.81m/s^{2} )=4709N[/tex]

The force applied to an item in curved motion that is pointed toward the axis of rotation or the centre of curvature is known as a centripetal force. Newtons are used to measure centripetal force. The direction of the centripetal force is always perpendicular to the direction in which the item is moving.

However, the centripetal force, which keeps the car turning in a circle, is also related to the frictional force:

[tex]F_{f} =F_{c} =m \frac{v^{2} }{r}[/tex]

Where v is the fastest the car can travel while still in the turn. We may get the value of v by rearranging the formula and substituting r=88.0 m:

[tex]v=\sqrt{} \frac{Fx_{r} }{m} = \sqrt{}\frac{(4709N)(88..0m)}{1200kg}[/tex]

18.6 m/s

Learn more about static friction:

https://brainly.com/question/25534663

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