Answer :
The boxes acceleration up the ramp is approximately 0.109 m/s².
The forces acting on the box can be resolved into two components: one parallel to the incline (F_parallel) and one perpendicular to the incline (F_perpendicular).
Given:
Mass of the box (m) = 40.7 kg
Coefficient of kinetic friction (μ) = 0.17
The force pulling the box (F_p) = 173 N
Incline angle (θ₁) = 14.5°
Force angle (θ₂) = 25.7°
First, we need to calculate the components of the force pulling the box:
F_parallel = F_p * sin(θ₂)
F_perpendicular = F_p * cos(θ₂)
Next, let's calculate the force of friction:
F_friction = μ * (mass of the box) * g, where g is the acceleration due to gravity (approximately 9.8 m/s²)
The force component parallel to the incline is opposed by the force of friction, so:
Net force parallel to the incline (F_net_parallel) = F_parallel - F_friction
Now, we can calculate the acceleration using Newton's second law:
F_net_parallel = (mass of the box) * acceleration
Rearranging the equation, we get:
acceleration = F_net_parallel / (mass of the box)
Now we can substitute the values into the equations and calculate the acceleration
F_parallel = 173 N * sin(25.7°) ≈ 73.88 N
F_perpendicular = 173 N * cos(25.7°) ≈ 154.37 N
F_friction = 0.17 * (40.7 kg) * 9.8 m/s² ≈ 69.44 N
F_net_parallel = F_parallel - F_friction ≈ 73.88 N - 69.44 N ≈ 4.44 N
acceleration = (4.44 N) / (40.7 kg) ≈ 0.109 m/s²
Therefore, the box's acceleration up the ramp is approximately 0.109 m/s².
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