Question

A car drives over a rounded hill.

A) What is the fastest the car can go, in m/s, without the tires coming off of the road?

B) If the mass of the car is 1500 kg, what is the net force on the car when it is at the top of the hill when it travels at a speed of 5 m/s?

Answer 1

Part A

To start this problem off, the first thing you should do is draw a free body diagram (FBD). Since the car is going in a circle on this turn, we know that the car is going to feel a centripetal force towards the center of the circle. What provides this centripetal force is the friction force of the road. So the friction force of the road is

F(fr)=Nu

- F(fr)=friction force in Newtons
- N is the normal force in newtons
- u is the coefficient of static friction

Since the car is sitting on a horizontal surface, the normal force will equal the magnitude of the weight force so

F(fr)=mgu

m is mass and g is acceleration due to gravity

Like I just said, this friction force must equal the centripetal force to keep the car turning around the bend. So

F(c)=mv^2/r

- F(c) is the centripetal force in newtons
- m is the mass of the object in kilograms
- v is the speed at which the object goes around the curve
- r is the radius of the circle

We now set the equations equal to each other

mv^2/r=mgu

m cancels from both sides of the equation so

v^2/r=gu

multiply both sides by r

v^2=gru

take the square root of both sides and you have an equation solving for velocity

v=(gru)^1/2 m/s since static friction, radius of hill is unknown..you can plug in g as 9.81.

Part B

F net force= mv^2/r (centripetal force)+ mgu(friction force)

                = 1500*(5)^2/r + 1500*9.81*u   where r is radius and u is static friction