Respuesta :
Answer:
d = 0.544 m
Explanation:
To solve this problem we must work in two parts: one when the surface has no friction and the other when the surface has friction
Let's start with the part without rubbing, let's find the speed that the box reaches., For this we use the conservation of mechanical energy in two points: maximum compression and when the box is free (spring without compression)
Initial, maximum compression
Em₀ = Ke = ½ k x²
Final, free box without compressing the spring
[tex]Em_{f}[/tex] = K = ½ m v²
Emo = [tex]Em_{f}[/tex]
½ k x² = ½ m v²
v = √ (k / m) x
Let's reduce the SI units measures
x = 20 cm (1m / 100cm) = 0.20 m
v = √ (100 / 2.5) 0.20
v = 1,265 m / s
Let's work the second part, where there is friction. In this part the work of the friction force is equal to the change of mechanical energy
[tex]W_{fr}[/tex] = ΔEm = [tex]Em_{f}[/tex] - Em₀
[tex]W_{fr}[/tex] = - fr d
Final point. Stopped box
[tex]Em_{f}[/tex] = 0
Starting point, starting the rough surface
Em₀ = K = ½ m v²
With Newton's second law we find the force of friction
fr = μ N
N-W = 0
N = W = mg
fr = μ mg
-μ m g d = 0 - ½ m v²
d = ½ v² / (μ g)
Let's calculate
d = ½ 1,265² / (0.15 9.8)
d = 0.544 m
The distance traveled by the box along the rough surface before stopping is 0.54 m.
The given parameters;
- spring constant, k = 100 N/m
- extension of the spring, x = 20 cm = 0.2
- mass of the box, m = 2.5 kg
- coefficient of kinetic friction, μ = 0.15
The distance traveled by the box is determined by using work-energy theorem as shown below;
[tex]\mu_k Fd = \frac{1}{2}kx^2\\\\d = \frac{kx^2}{2\mu_k F} \\\\d = \frac{kx^2}{2\mu_k \times mg} \\\\d = \frac{100(0.2^2)}{2\times 0.15 \times 2.5 \times 9.8} \\\\d = 0.54 \ m[/tex]
Thus, the distance traveled by the box along the rough surface before stopping is 0.54 m.
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