Respuesta :
Answer:
[tex]5.94\ \text{m/s}[/tex]
[tex]1.7[/tex]
[tex]0.577[/tex]
Explanation:
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
[tex]\theta[/tex] = Angle of slope = [tex]30^{\circ}[/tex]
v = Velocity of child at the bottom of the slide
[tex]\mu_k[/tex] = Coefficient of kinetic friction
[tex]\mu_s[/tex] = Coefficient of static friction
h = Height of slope = 1.8 m
The energy balance of the system is given by
[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}[/tex]
The speed of the child at the bottom of the slide is [tex]5.94\ \text{m/s}[/tex]
Length of the slide is given by
[tex]l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}[/tex]
[tex]v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}[/tex]
The force energy balance of the system is given by
[tex]mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73[/tex]
The coefficient of kinetic friction is [tex]1.7[/tex].
For static friction
[tex]\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577[/tex]
So, the minimum possible value for the coefficient of static friction is [tex]0.577[/tex].
The speed of the child depends on the friction coefficient between the
surface of the slide and the surface on which the child seats.
- Part A: The speed of the child when he reaches the bottom of the side is approximately 5.943 m/s.
- Part B: The coefficient of kinetic friction is approximately 0.433.
- Part C: the minimum possible value of the coefficient of static friction for the child and the surface of the slide is approximately 0.557.
Reasons:
The height of the slide, h = 1.8 m.
Angle of elevation of the slide = 30°
Part A:
Nature of the surface of the slide = Frictionless
The potential energy of the slide is converted into kinetic energy
Potential energy = m·g·h
Kinetic energy, K.E. = [tex]\mathbf{\dfrac{1}{2} \cdot m \cdot v^2}[/tex]
Where:
g = The acceleration due to gravity
Therefore;
[tex]m \cdot g \cdot h = \dfrac{1}{2} \cdot m \cdot v^2[/tex]
[tex]v = \sqrt{2 \cdot g \cdot h} = \sqrt{2 \times 9.81 \times 1.8} \approx 5.943[/tex]
The speed of the child when he reaches the bottom of the side, v ≈ 5.943 m/s.
Part B: The speed at the bottom is half the value calculated in part A, therefore;
[tex]v_2 \approx \dfrac{5.943}{2} = 2.9715[/tex]
Energy used to do work on friction, ΔW = 0.5·m·v² - 0.5·m·v₂²
Which gives;
ΔW = 0.5×m × (5.943² - 2.9715²) = 13.245·m
Work by friction, ΔW = [tex]\mu_k[/tex]·m·g·cos(θ)×L
Where;
Box On An Incline
L = The length of the slide = [tex]\dfrac{h}{sin(\theta)}[/tex]
Therefore;
[tex]\Delta W = \mathbf{\mu_k \cdot m \cdot g \cdot cos (\theta) \times \dfrac{h}{sin(\theta)}}[/tex]
Which gives;
[tex]\Delta W = \mu_k \cdot m \times 9.81 \times cos (30 ^{\circ}) \times \dfrac{1.8}{sin(30 ^{\circ})} = \mathbf{13.245 \cdot m}[/tex]
[tex]\mu_k = \dfrac{ 13.245 \cdot m}{m \times 9.81 \times cos (30 ^{\circ}) \times \dfrac{1.8}{sin(30 ^{\circ})} } \approx 0.433[/tex]
The coefficient of kinetic friction, [tex]\mu_k[/tex] ≈ 0.433
Part C: When the child sits and remain at rest, we have;
Force of friction = The weight of the child acting along the plane
Therefore;
[tex]\mu_s[/tex]·m·g·cos(θ) ≥ m·g·sin(θ)
[tex]\mu_s \geq \dfrac{m \cdot g \cdot sin(\theta)}{m \cdot g \cdot cos(\theta)} = \mathbf{tan(\theta)}[/tex]
[tex]\mu_s[/tex] ≥ tan(30°) = [tex]\dfrac{\sqrt{3} }{3}[/tex] ≈ 0.557
Therefore;
The minimum possible value of the coefficient of static friction for the child and the surface of the slide, [tex]\mu_s[/tex] ≈ 0.557.
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