Question

A car of mass 2000 kg is moving with a speed of 10 metre per second on a circular path of radius 20 meters on a level road what must be the frictional force between the car and the road so that the car does not slip.​

Answer 1

The frictional force is =1178.75N

Explan.

The mass of the car is m=1500kg

The radius of the track is r=20m

The speed of the car is v=12.5ms−1

The frictional force is equal to the centripetal force to prevent slipping

Fr=mv2r=1500⋅12.5220=11718.75N

Answer 2

Answer:

Thus, the required frictional force between the car and the road so that the car does not slip is $10^{4} N$.​

Explanation:

Concept:

The opposing force that is produced when two surfaces attempt to move in the same direction or in opposite directions is known as frictional force. A frictional force is primarily intended to produce resistance to the mobility of one surface across another surface.

Given:

Mass of car $m$ $=2000 kg$

Speed $v$ $=10m/s$

Radius of the path $r$$=20 m$

To find:

To prevent the car from slipping, we must determine the frictional force between the road and the vehicle.

Solution:

Frictional force will function in a manner that generates the required centripetal force, preventing sliding. I.e., the centre of the circular curvature will be affected by frictional force.

$f=\frac{mv^{2} }{r}$

Here, $f$ is magnitude of frictional force.

$f=\frac{2000*10^{2}}{20}$

$f=10^{4} N$

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