Question

derive an expression of maximum velocity of a car on a level road

Answer 1

Answer:

The maximum velocity of a car on a level road will be $v\leq\sqrt{\mu g r}$.

Explanation:

The normal reaction acting on the car on a level road is given as,

$N=mg$    (1)

Where,

N=normal reaction acting on the car

m=mass of the car

g=acceleration due to gravity=10m/s²

Here the static friction will provide the necessary centripetal acceleration.  Which is given as,

$f\leq\mu N$     (2)

f=static friction

μ=coefficient of friction

N=normal reaction

Since the static friction provides necessary centripetal acceleration it is also calculated as,

$f=\frac{mv^2}{r}$     (3)

v=velocity of the car

m=mass of the car

r=radius of the car

By substituting equations (1) and (3) in equation (2) we get;

$\frac{mv^2}{r}\leq\mu \times mg$

$v^2\leq \mu gr$

$v\leq\sqrt{\mu g r}$

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Answer 2

Answer:

The expression of maximum velocity of a car on a level road is $v\leq\sqrt{\mu g r}$

Explanation:

As per the data given in the question,

we have,

Acting Force on the car is normal reaction Force,

N=mg........(i)

where N=normal reaction on car; m=mass of car; g= gravitational acceleration=10m/s²

Also, static friction will be provided here to the car so,

$f\leq\mu N$.........(ii)

where f=static friction; μ=friction coefficient; N=normal reaction

As, Static friction provides the centripetal acceleration,so

$f=\frac{mv^2}{r}$..........(iii)

where v=velocity of car; m=mass of car; r=radius of car

From (i) (ii) (iii) we get,

$\frac{mv^2}{r}\leq\mu \times mgv^2\leq \mu grv\leq\sqrt{\mu g r}$

The expression of maximum velocity of a car on a level road is $v\leq\sqrt{\mu g r}$

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