A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coefficient of friction present between the roadway and the wheels of the truck

The coefficient of friction present between the roadway and the wheels of the truck is 0.833.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

Therefore, frictional force,

Now, frictional force = centripetal force

Plug in the given values and solve for 'μ'. This gives,

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2