(a) The acceleration is (6.05 - 4.73 × µ) m/s²
(b) Friction Force is f = 23.2 µ N, (c) Minimum friction coefficient is 0.80
(a) Acceleration:
The hollow sphere will experience two forces:
gravitational force (mg) that pulls it down the slope, and friction force (f) that acts opposite to the direction of motion. The component of gravitational force down the slope is mgsinθ and that perpendicular to the slope is mgcosθ.
The net force F on the hollow sphere down the slope is given by:
F = mgsinθ - f.
The magnitude of the force of friction f is:
f = µmgcosθ. Substitute this value of f in the expression for F to get: F = mgsinθ - µmgcosθ.
The acceleration a of the hollow sphere down the slope is given by the equation: a = F/m. Substitute the value of F to get: a = (mgsinθ - µmgcosθ)/m = g(sinθ - µcosθ). This is the acceleration of the hollow sphere down the slope.
Therefore, acceleration is a = g(sinθ - µcosθ)
Substituting the given values, a = 9.81 × (sin 39.0° - µcos 39.0°)
a = 6.05 - 4.73 µ
a = (6.05 - 4.73 × µ) m/s²
(b) Friction force:
The friction force f acting on the hollow sphere is given by: f = µmgcosθ. Substitute the values of m, g, θ, and µ to get: f = µ × 3.00 kg × 9.81 m/s² × cos 39.0°. Simplify to get: f = 23.2 µ N.
Therefore, friction force is f = 23.2 µ N
(c) Minimum friction coefficient:
The minimum friction coefficient µ_min needed to prevent slipping is given by the expression: µ_min = tanθ. Substitute the value of θ to get: µ_min = tan 39.0°. Simplify to get: µ_min = 0.80.
Therefore, the minimum friction coefficient is µ_min = 0.80.
If the mass were doubled to 6.00 kg, then the acceleration and the friction force will both double. The new acceleration will be: a' = 2a = 2g(sinθ - µcosθ) = 2(9.81 m/s²)(sin 39.0° - µcos 39.0°) = 2(6.05 - 4.73 µ) m/s².
Therefore, acceleration is a' = 2(6.05 - 4.73 µ) m/s²
The new friction force will be: f' = 2f = 2µmgcosθ = 2 × 23.2 µ N = 46.4 µ N.
Therefore, friction force is f' = 46.4 µ N.
The minimum friction coefficient will remain the same at µ_min = 0.80.
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