Question

A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the friction force required to keep the car from sliding?

Answer 1

Answer:

4500 N

Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

a = v^2/r

where v is the velocity of the body and r is the radius of the circumference:

Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

fs = μN = μmg

The car will not slide if     f = fs,   i.e.

fs = μmg =  m v^2/r

That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration

fs = (1000 kg)  * (30m/s)^2 / (200 m) = 4500 N

Answer 2

Explanation:

It is known that relation between force, mass, velocity and radius is as follows.

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

As diameter is given as 0.20 km. So, radius is $\frac{diameter}{2}$. Hence, radius will be equal to 0.10 km.

As, 1 km = 1000 m. Therefore, 0.10 km = 100 m.

Also, it is given that mass is 1000 kg and velocity is 30 m/s. Hence, calculate the force required to keep the car from sliding will be as follows.

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

                       = $\frac{1000 kg \times (30)^{2}}{100 m}$

                       = 9000 N

Thus, we can conclude that the magnitude of the friction force required to keep the car from sliding is 9000 N.