Question

A 13,900 N car traveling at 40.0 km/h rounds a curve of radius 1.80 ✕ 102 m.

Answer 1

a) $0.68 m/s^2$

b) 964.5 N

c) 0.069

Explanation:

a)

When an object is moving in a circular motion, the direction of its velocity is changing - therefore, it has an acceleration towards the center of the circle, called centripetal acceleration.

The magnitude of the centripetal acceleration is given by

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

where

v is the speed of the object

r is the radius of the circle

For the car in this problem:

v = 40.0 km/h = 11.1 m/s is the speed

r = 180 m is the radius of the circle

Substituting, we find the acceleration:

$a=\frac{11.1^2}{180}=0.68 m/s^2$

b)

The centripetal force is the force that keeps the object along its circular motion. It also acts towards the center of the circle, and it is given by

$F=ma$

where

m is the mass of the object

a is the centripetal acceleration

Here the weight of the car is

$W=mg=13,900 N$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

So the mass is

$m=\frac{W}{g}=\frac{13,900}{9.8}=1418.4 kg$

Therefore, the centripetal force is

$F=(1418.4)(0.68)=964.5 N$

c)

In this case, the force of static friction between the tires and the road provides the required centripetal force to keep the car in circular motion. This force is given by:

$F_f=\mu mg$

where

$\mu$ is the coefficient of friction

Equating the frictional force to the centripetal force,

$\mu mg=ma$

So we get:

$\mu=\frac{a}{g}$

And substitutng:

$a=0.68 m/s^2$ (centripetal acceleration)

$g=9.8 m/s^2$

We find:

$\mu=\frac{0.68}{9.8}=0.069$