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Sagot :
Answer:
i) a₁ = -g (sin θ + μ cos θ), x = v₀² / 2a₁
ii) W = mg L sin θ , iii) Wₙ = 0
iv) W = - μ m g L cos θ x
Explanation:
With a drawing this exercise would be clearer, I understand that you have a block on a ramp and it is subjected to some force that makes it rise, for example the tension created by a descending block.
The movement is that when the system is released, the tension forces are greater than the friction and the component of the weight and therefore the block rises up the ramp
At some point the tension must become zero, when the hanging block reaches the ground, as the block has a velocity it rises with a negative acceleration to a point and stops where the friction force and the weight component would be in equilibrium along the way. along the plane
i) Let's use Newton's second law
the reference system is with the x axis parallel to the ramp
Axis y
N - W cos θ = 0
X axis
T - W sin θ - fr = ma
the friction force is
fr = μ N
fr = μ mg cos θ
we substitute
T - m g sin sin θ - μ mg cos θ = m a
a = T / m - g (sin θ + μ cos θ)
With this acceleration we can find the height that the block reaches, this implies that at some point the tension becomes zero, possibly when a hanging block reaches the floor.
T = 0
a₁ = -g (sin θ + μ cos θ)
v² = v₀² - 2a1 x
v = 0 at the highest point
x = v₀² / 2a₁
ii) the work of the gravitational force is
W = F .d
W = mg sin θ L
iii) the work of the normal force
the force has 90º with respect to the displacement so cos 90 = 0
Wₙ = 0
iv) friction force work
friction force always opposes displacement
W = - fr d
W = - μ m g cos θ L
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