Respuesta :
Answer:
[tex]a = 5.1\ m/s^2[/tex]
Explanation:
Given,
The angle of the slide=[tex] 42^\circ[/tex]
The mass of the child is= m
coefficient of friction = 0.20
when she slides down now apply Newton's law
[tex]ma =mg \sin\theta - f[/tex]
[tex]ma = mg\sin \theta -\mu mg \cos \theta[/tex]
therefore the acceleration
[tex]a=g \sin\theta -\mu gcos\theta [/tex]
[tex]a=g[\sin \theta -\mu \cos \theta][/tex]
[tex]a=9.8\times [\sin 42^\circ -0.2\times \cos 42^\circ][/tex]
[tex]a = 5.1\ m/s^2[/tex]
hence, the magnitude of acceleration during her sliding is equal to [tex]a = 5.1\ m/s^2[/tex]
The magnitude of the acceleration of the child is 6.54 m/s²
Acceleration:
The angle of incline is θ = 42°
The coefficient of kinetic friction, μ = 0.20
The weight of the child splints into two components:
(i) perpendicular downward the inclined surface: mgcosθ
(ii) along the surface of the incline: mgsinθ
The frictional force acts opposite to the motion of the child down the inclined surface given by:
f = μmgcosθ
So the equation of motion is given by:
ma = mgsinθ - μmgcosθ
a = g(sinθ - μcosθ)
a = 9.8(sin42 - 0.2×cos42)
a = 6.54 m/s²
Learn more about acceleration:
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