Physics, 22.01.2020 01:31 Officialavm

A14000n car traveling at 25m/s rounds a curve of radius 200m. find the following: a. the centripetal acceleration of the car.

b. the force that maintains centripetal acceleration.

c. the minimum coefficient of static friction between the tires and road that will allow the car to round the curve safely.

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Answer from: superbatman9193

a. The centripetal acceleration of the car is equal to 3.125 m/s^2

b. The force that maintains centripetal acceleration is equal to 1400 kilograms.

c. The minimum coefficient of static friction between the car tires and road is equal to 0.31.

Given the following data:

Weight of car = 14000 Newton.Velocity of car = 25 m/s.Radius of curve = 200 meters.

Scientific data:

Acceleration due to gravity = 10 m/s^2

a. To determine the centripetal acceleration of the car:

Mathematically, the centripetal acceleration of an object is given by the formula:

$A_c = \frac{V^2}{r}$

Where;

is the centripetal acceleration.r is the radius of the circular track.V is the velocity of an object.

Substituting the given parameters into the formula, we have;

$A_c = \frac{25^2}{200}\\\\A_c = \frac{625}{200}\\\\A_c = 3.125 \;m/s^2$

b. To determine the force that maintains centripetal acceleration:

First of all, we would find the mass of the car.

$Weight = mg\\\\14000 = mass \times 10\\\\Mass =\frac{14000}{10}$

Mass = 1400 kilograms.

Note: The force that maintains centripetal acceleration is a centripetal force.

Mathematically, centripetal force is given by the formula:

$F_c = \frac{mv^2}{r}\\\\F_c = \frac{1400 \times 25^2}{200}\\\\F_c = \frac{1400 \times 625}{200}\\\\F_c = 4375\; Newton$

c. To determine the minimum coefficient of static friction between the car tires and road:

$F_c = uF_N\\\\F_c = umg\\\\4375=u(1400\times10)\\\\4375=14000u\\\\u=\frac{4375}{14000}$

Minimum coefficient of static friction = 0.31

Read more on centripetal acceleration here: link

Answer from: carmenala2

Explanation:

Given

Weight of car $W=14,000\ N$

mass of car $m=\frac{14,000}{9.8}=1428.57\ N$

velocity of car $v=25\ m/s$

radius $r=200\ m$

(a)Centripetal acceleration is given by

$a_c=\frac{v^2}{r}$

$a_c=\frac{25^2}{200}$

$a_c=3.125\ m$

(b)Force that provide centripetal acceleration

$F=F_c=\frac{mv^2}{r}$

$F=\frac{1428.57\times 25^2}{200}$

$F=4464.285\ N$

(c)Friction force between car and tires is given by

$=\mu N$

where $\mu$ =coefficient of static friction

N=normal reaction

Centripetal force will balance the friction force

F_c=F_r

$4464.285=\mu \times 1428.57\times 9.8$

$\mu =0.318$

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