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saw5 [17]
3 years ago
7

Queremos diseñar un montacargas que pueda subir con una rapidez de 12 km/h una mas 700 kg hasta 40 m de altura en un minuto. Cal

cula: a) El trabajo que realiza en ese recorrido. b) La potencia de motor que necesita. c) la energía cinética d) energía potencial d) la energía mecánica
Physics
1 answer:
Mariulka [41]3 years ago
8 0

Answer:

a) El trabajo realizado es de 274,680 J

b) La potencia de la carretilla elevadora es de 4578 Watts.

c) La energía cinética del montacargas es de 3.888.\overline 8 J

d) La energía potencial del montacargas es de 274.680 Joules.

e) La energía mecánica de la carretilla elevadora 278,568.\overline 8 J

Explanation:

a) Los parámetros dados son;

La velocidad de la carretilla elevadora, v = 12 km / h = 10/3 m / s

La masa que debe levantar la carretilla elevadora, m = 700 kg

La altura a la que se levantará la masa, h = 40 m

El trabajo realizado, W = Fuerza, F × Distancia, h

 La fuerza, F aplicada = El peso de la carga = Masa, m × Gravedad, g

Donde 'g' es la aceleración debida a la gravedad ≈ 9,81 m / s²

∴ Trabajo realizado, W = 700 kg × 9,81 m / s² × 40 m = 274,680 J

b) El tiempo que se tarda en subir 40 m = 1 minuto = 60 segundos

∴ Potencia = Trabajo / tiempo

Por lo tanto, la potencia del montacargas, P = 274,680 J / (60 s) = 4578 Watts

c) Energía cinética, K.E. = 1/2 · m · v²

La energía cinética de la carretilla elevadora, K.E. se da como sigue;

Carretilla elevadora K.E. = 1/2 × 700 kg × (10/3 m / s) ² = 3.888.\overline 8 J

d) La energía potencial del montacargas a 40 m, P.E. = m · g · h

∴ P.E. = 700 kg × 9,81 m / s² × 40 m = 274,680 Julios

e) La energía mecánica, M.E. = P.E. + K.E.

∴ M.E. = 3.888.\overline 8 J + 274,680 J = 278,568.\overline 8 J

La energía mecánica de la carretilla elevadora, M.E.= 278,568.\overline 8 J.

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