William C. answered 10/17/23
Experienced Tutor Specializing in Chemistry, Math, and Physics
m = 43.0 kg, θ = 25°
(a) The x-component of F must counter-balance the static frictional force fₛ in order to get the block moving.
The y-component of F impacts fₛ by decreasing the normal force on the block.
Fₓ = F cos θ and Fᵧ = F sin θ
Fₙ = mg – Fᵧ = mg – F sin θ
fₛ = μₛ Fₙ = μₛ( mg – F sin θ)
The minimum force needed to get the block moving is given by Fₓ > fₛ which means that
F cos θ > μₛ( mg – F sin θ) = μₛmg – μₛ F sin θ
F(cos θ + μₛ sin θ) > μₛmg
F > (μₛmg)/(cos θ + μₛ sin θ)
Plugging in numbers (m = 43.0 kg, θ = 25°, g = 9.8 m/s²)
F > (421.4μₛ)/(0.906 + 0.423μₛ)
where μₛ is coefficient of static friction
(b) Fnet = F cos θ – μₖ( mg – F sin θ) = F(cos θ + μₖ sin θ) – μₖmg
a = Fnet/m = F(cos θ + μₖ sin θ)/m – μₖg
Plugging in numbers (m = 43.0 kg, θ = 25°, g = 9.8 m/s²)
a = (0.0211 + 0.0098 μₖ)F – 9.8 μₖ
where μₖ is coefficient of kinetic friction