Question

The maximum speed with which a 945-kg car makes a 180-degree turn is 10.0 m/s. The

radius of the circle through which the car is turning is 25.0 m. Determine the force of

friction and the coefficient of friction acting upon the car.​

Answer 1

$\sf{\huge{\bold{\underline{ Solution }}}}$

Given :-

• The mass of the car is 945 kg
• Angle formed by car is 180°
• Speed of the car is 10m/s
• Radius of the car is 25 m

To Find :-

• We have to determine the force of friction and the coffiecients of friction acting upon the car?

Let's Begin :-

Here,

• Normal force ( R) is acting in upward direction whereas Force of gravity is acting downwards

Therefore,

Force of gravity acting in downward direction

$\bold{\pink{ = mg }}$

$\sf{ = 945 }{\sf{\times{9.8}}}$

$\bold{ = 9261 N }$

Now,

According to the question ,

• Vertical components of force balance each other as they are acting opposite to each other

Therefore, It concludes that

• Normal force = Force of gravity

That is ,

$\sf{ Fn = Fg = 9216 N }$

Thus, Both the vertical forces are equal

Now,

We have to determine the force of friction

Here, Force of friction is acting in horizontal direction . Also, Angle formed by the car is 180° . So,

• Force of friction = Net Force on the car

That is,

$\sf{ F = m }{\sf{\times{\dfrac{ v^{2}}{R}}}}$

• R is the radius of circle and v² is the speed of the car

Subsitute the required values,

$\sf{ F = 945 }{\sf{\times{\dfrac{ (10)^{2}}{25}}}}$

$\sf{ F = 945 }{\sf{\times{\dfrac{100}{25}}}}$

$\sf{ F = 945}{\sf{\times{ 4 }}}$

$\bold{ F = 3780 N }$

Here, we got the force of Friction acting on the car is 3780 N

Now

We have to determine the friction coffiecients

We know that,

• Friction coffiecient (μ) = Force of friction /Normal force

That is,

$\sf{\mu{ = }}{\sf{\dfrac{ 3780}{9261}}}$

$\bold{\mu{ = 0.408 }}$

Hence, The force of friction and friction coffiecients are 3780 N and 0.408 .

[ Note :- Please refer the above attachment ]