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.

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.

Encuentra más sobre trabajo aqui:

Espero que te sea de utilidad!  

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