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Elza [17]
2 years ago
15

Un bloque de 20kg de masa se desplaza horizontalmente en la dirección de eje X por acción de una fuerza horizontal variable F =

6x, donde F se mide en Newton y "x" en metros. Si se desprecia el rozamiento determine: a) El trabajo realizado por esta fuerza mientras el bloque se mueve desde la posición x=+10m hasta la posición x=+20m b) La rapidez del bloque en la posición x=+20m, si su rapidez en la posición x=+10m fue √10 m/s.
Physics
1 answer:
zmey [24]2 years ago
6 0

Answer:

a) El trabajo realizado por esta fuerza mientras el bloque se mueve desde la posición x = + 10 m hasta la posición x = + 20 m es 900 joules.

b) La rapidez del bloque en la posición x = + 20 metros es aproximadamente 5.701 metros por segundo.

Explanation:

a) El trabajo expermentado por el bloque (W), medido en joules, es definida por la siguiente ecuación integral:

W = \int\limits^{x_{max}}_ {x_{min}} F(x) \, dx (1)

Donde:

x_{min}, x_{max} - Posiciones mínima y máxima del bloque, medidos en metros.

F(x) - Fuerza horizontal aplicada al bloque, medida en newtons.

Si conocemos que F(x) = 6\cdot x, x_{min} = 10\,m y x_{max} = 20\,m, entonces el trabajo realizado por esta fuerza es:

W = \int\limits^{20\,m}_{10\,m} {6\cdot x} \, dx (2)

W = 6\int\limits^{20\,m}_{10\,m} x\, dx

W = 3\cdot x^{2}\left|\limits_{10\,m}^{20\,m}

W = 3\cdot [(20\,m)^{2}-(10\,m)^{2}]

W = 900\,J

El trabajo realizado por esta fuerza mientras el bloque se mueve desde la posición x = + 10 m hasta la posición x = + 20 m es 900 joules.

b) La rapidez final del bloque se determina mediante de Teorema del Trabajo y la Energía, es decir:

W = K_{f}-K_{o} (3)

Donde son K_{o}, K_{f} las energías cinéticas traslacionales inicial y final, medidos en joules.

Al aplicar la definición de energía cinética traslacional, expandimos y simplificamos la ecuación como sigue:

W = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (4)

Donde:

m - Masa del bloque, medido en kilogramos.

v_{o}, v_{f} - Rapideces inicial y final del bloque, medidos en metros por segundo.

\frac{2\cdot W}{m} = v_{f}^{2}-v_{o}^{2}

v_{f} = \sqrt{\frac{2\cdot W}{m}+v_{o}^{2}}

Si conocemos que W = 900\,J, m = 20\,kg y v_{o} = \sqrt{10}\,\frac{m}{s}, entonces la rapidez final del bloque es:

v_{f} = \sqrt{\frac{900\,J}{2\cdot (20\,kg)}+10\,\frac{m^{2}}{s^{2}}  }

v_{f} \approx 5.701\,\frac{m}{s}

La rapidez del bloque en la posición x = + 20 metros es aproximadamente 5.701 metros por segundo.

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