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.
<h3>What is frictional force?</h3>
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.
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