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Sagot :
Answer:
a = 1,538 m / s²
Explanation:
Let's use Newton's second law, let's set a reference system where the x-axis is parallel to the sloping floor of the truck and the positive direction is in the direction of movement of the trough, for this case the weight is the only force to decompose
sin θ = Wₓx / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Y axis
N -W_y = 0
N = mg cos θ
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
There are values of the friction coefficient (μ_s) one for when the movement has not started and it takes a smaller value for when the bodies are moving.
In this case we first find the angle for which the movement begins, in this part we use the static coefficient and the acceleration is zero
Wₓ - μ_s N = 0
m g sin θ = μ_s mg cos θ
tan θ = μ_s
θ = tan⁻¹ μ_s
we calculate
θ = tan⁻¹ 0.8
θ = 38.7º
For this angle, how the trough begins to move, the coefficient is reduced to the dynamics coefficient (μ_k) and the acceleration is different from zero.
we substitute
mg sin θ - μ_k mg cos θ = m a
a = g (sin θ - μ_k cos θ)
let's calculate
a = 9.8 (sin 38.7 - 0.6 cos 38.7)
a = 1,538 m / s²
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