1 A steel ball of mass 5 kg is released from rest at a height of 3 m above the ground. It rolls down the frictionless section AB and then moves up the rough inclined section BC before coming to a stop at point C, which is at height of 2 m above the ground as in shown in the diagram.

2.1.1 Hence, use the law stated above to calculate the speed of the ball when it reaches point B [4]

The kinetic frictional force between the ball and surface of section BC is 10N

2.1.2 use energy principles to calculate the angle of inclination, of section BC. [6]

Question

contestada

Respuesta :

sp10925 sp10925 · Answer 1 · 2022-05-17T11:02:05+02:00

The speed of the ball when it reached the point B is 7.7 m/s. The angle of inclination of section BC is 11.54⁰.

The force between the two mating surfaces and having the relative motion is called the frictional force.

The speed of ball at point B is equal to the relation

v = √(2gh)

Substitute the values of height h = 3m and acceleration due to gravity g =9.8 m/s², we get the speed at B

v =√(2 x 9.8 x 3)

v =7.7 m/s

Thus the speed of ball when it reaches the point B is 7.7 m/s.

Given is the mass of steel ball is 5 kg and the frictional force is 10N.

The frictional force is equal to the product of coefficient of friction and the normal force on the steel ball \.

f = μN

f =μmg

Substitute the values into the equation, we get the coefficient of friction.

10 = μ x 5 x 9.8

μ = 0.2041

Angle of inclination for the section BC is

θ = tan⁻¹(0.2041) = 11.54⁰

Thus, the angle of inclination of section BC is 11.54⁰.

Learn more about frictional force.

https://brainly.com/question/14662717

#SPJ1