El cuerpo de 2kg se suelta de A recorriendo el plano inclinado 60° con la horizontal y de superficie lisa hasta el punto B. Luego recorre el tramo rugoso BC deteniéndose en C. Hallar “L” si se sabe que μ= 0,6.
Answers
El valor de L del tramo rugoso B a C con coeficiente de fricción igual a 0,6, recorrido por el cuerpo de 2 kg cuando se suelta del punto A es 5.00 m.
El valor de L se puede calcular a partir de la definición de trabajo:
[tex]W = F_{\mu}*d[/tex] (1)
En donde:
[tex]F_{\mu}[/tex]: es la fuerza aplicada sobre el cuerpo en el tramo de B a C = fuerza de roce = -μN (el signo menos se debe a que está en dirección opuesta a la del movimiento)
μ: es el coeficiente de roce = 0,6
N: es la normal = mg
m: es la masa del cuerpo = 2 kg
g: es la aceleración debida a la gravedad = 9,81 m/s²
d: es la distancia = L =?
Por otra parte, el trabajo también se define como la diferencia de energía mecánica entre los puntos B y C.
[tex] W = E_{C} - E_{B} [/tex] (2)
Al igualar la ecuación (1) con la (2) tenemos:
[tex]E_{C} - E_{B} = F_{\mu}*d[/tex] (3)
En el punto C, la energía es cero (0) dado que el cuerpo se detiene y en el punto B la energía que tiene el objeto es la cinética:
[tex] E_{B} = \frac{1}{2}mv_{B}^{2} [/tex]
En donde:
[tex]v_{B}[/tex]: es la velocidad del cuerpo en el punto B
La velocidad del cuerpo en el punto B se puede calcular mediante conservación de energía entre los puntos A y B:
[tex] E_{A} = E_{B} [/tex]
En el punto A, el cuerpo tiene energía potencial gravitacional, por lo tanto:
[tex] mgh = \frac{1}{2}mv_{B}^{2} [/tex] (4)
En donde:
h: es la altura = 3 m
Entonces, la velocidad en el punto B es (eq 4):
[tex] v_{B} = \sqrt{\frac{2mgh}{m}} = \sqrt{2*9,81 m/s^{2}*3 m} = 7,67 m/s [/tex]
Ahora, la ecuación (3) queda como sigue:
[tex] 0 - \frac{1}{2}mv_{B}^{2} = -\mu N*L [/tex]
[tex] \frac{1}{2}2 kg*(7.67 m/s)^{2} = 0,6*2 kg*9,81 m/s^{2}*L [/tex]
Resolviendo para L, tenemos:
[tex] L = \frac{\frac{1}{2}2 kg*(7.67 m/s)^{2}}{0,6*2 kg*9,81 m/s^{2}} = 5.00 m [/tex]
Por lo tanto, el valor de L es 5.00 m.
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