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Block on inclined plane with friction

A block of m = 5 kg is on an inclined plane at θ = 30°.\nThe kinetic friction coefficient is μ_d = 0.2. Calculate the acceleration of the sliding block.
Given data
m = 5 kgθ = 30°μ_d = 0.2g = 9.81 m/s²
Review the theory: Dinamica
Steps
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  1. What is the component of weight parallel to the inclined plane?
  2. What is the normal force exerted by the plane on the block?
  3. What is the kinetic friction force acting on the block?
  4. What is the acceleration of the block along the plane?
Full worked solution
  1. What is the component of weight parallel to the inclined plane?
    F∥=mgsin⁡θ=5⋅9.81⋅sin⁡30°F_\parallel = mg\sin\theta = 5 \cdot 9.81 \cdot \sin 30°
    F∥=mgsin⁡θ=5×9.81×sin⁡30∘=5×9.81×0.5=24.525 NF_\parallel = mg\sin\theta = 5 \times 9.81 \times \sin30^\circ = 5 \times 9.81 \times 0.5 = \mathbf{24.525\,N}. This is the component of weight parallel to the incline, pulling the block downward.
  2. What is the normal force exerted by the plane on the block?
    N=mgcos⁡θ=5⋅9.81⋅cos⁡30°N = mg\cos\theta = 5 \cdot 9.81 \cdot \cos 30°
    N=mgcos⁡θ=5×9.81×cos⁡30∘=5×9.81×0.866=42.48 NN = mg\cos\theta = 5 \times 9.81 \times \cos30^\circ = 5 \times 9.81 \times 0.866 = \mathbf{42.48\,N}. The normal force balances the perpendicular component of the weight.
  3. What is the kinetic friction force acting on the block?
    Fa=μd⋅N=0.2⋅42.48F_a = \mu_d \cdot N = 0.2 \cdot 42.48
    Fa=μdN=0.2×42.48=8.50 NF_a = \mu_d N = 0.2 \times 42.48 = \mathbf{8.50\,N} directed up the slope, opposing the motion. Kinetic friction always opposes the relative sliding.
  4. What is the acceleration of the block along the plane?
    a=F∥−Fam=24.525−8.4965a = \dfrac{F_\parallel - F_a}{m} = \dfrac{24.525 - 8.496}{5}
    a=(F∥−Fa)/m=(24.525−8.496)/5=16.029/5=3.21 m/s2a = (F_\parallel - F_a)/m = (24.525 - 8.496)/5 = 16.029/5 = \mathbf{3.21\,m/s^2}. Compact formula: a=g(sin⁡θ−μdcos⁡θ)=9.81(0.5−0.2⋅0.866)=9.81⋅0.327=3.21 m/s2a = g(\sin\theta - \mu_d\cos\theta) = 9.81(0.5 - 0.2\cdot0.866) = 9.81\cdot0.327 = 3.21\,\mathrm{m/s^2}.
Result:Block acceleration: a ≈ 3.21 m/s² down the plane. Compact formula: a=g(sin⁡θ−μdcos⁡θ)a = g(\sin\theta - \mu_d\cos\theta).
TheoryDinamicaFormula sheetPhysics 1

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