Problem:
A child weighing 30 kg is at the top of a 5-meter long slide with an incline of 30 degrees. The coefficient of kinetic friction between the child and the slide is 0.2. If the child starts from rest, what is their speed at the bottom of the slide?
Solution:
1. Identify the forces:
* Gravity (Weight): Acts vertically downwards, with a magnitude of mg, where m = 30 kg (mass) and g = 9.8 m/s² (acceleration due to gravity).
* Normal Force: Acts perpendicular to the slide, counteracting the component of gravity perpendicular to the slide.
* Friction: Acts parallel to the slide, opposing the motion, with a magnitude of μN, where μ = 0.2 (coefficient of kinetic friction) and N is the normal force.
2. Resolve forces:
* Parallel to the slide: The component of gravity parallel to the slide is mg sin(30°), which drives the child down.
* Perpendicular to the slide: The component of gravity perpendicular to the slide is mg cos(30°), which is balanced by the normal force (N = mg cos(30°)).
3. Apply Newton's Second Law:
* Net force = mass × acceleration
* The net force acting on the child is mg sin(30°) - μN = ma.
4. Solve for acceleration:
* Substitute N = mg cos(30°) into the equation:
* mg sin(30°) - μ(mg cos(30°)) = ma
* a = g(sin(30°) - μ cos(30°))
* a = 9.8 m/s² (0.5 - 0.2 × 0.866) ≈ 3.15 m/s²
5. Use kinematics to find velocity:
* We know the initial velocity (v₀ = 0 m/s), acceleration (a ≈ 3.15 m/s²), and distance (d = 5 m).
* Use the kinematic equation: v² = v₀² + 2ad
* v² = 0² + 2 × 3.15 m/s² × 5 m
* v² ≈ 31.5
* v ≈ √31.5 ≈ 5.61 m/s
Therefore, the child's speed at the bottom of the slide is approximately 5.61 m/s.
