Question

Calculate the work required so that a block of mass 5 kg initially at rest and placed on a horizontal frictionless surface, tied by a rope 20 cm long fixed at the other end, begins to rotate until it reaches a speed angular of 2 rps. Do you need to add more work to maintain this speed? Justify your answer by applying the principle of conservation of energy.

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Answer to a math question Calculate the work required so that a block of mass 5 kg initially at rest and placed on a horizontal frictionless surface, tied by a rope 20 cm long fixed at the other end, begins to rotate until it reaches a speed angular of 2 rps. Do you need to add more work to maintain this speed? Justify your answer by applying the principle of conservation of energy.

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Hester
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## Treball necessari per girar el bloc Podem analitzar aquest problema utilitzant els conceptes d'inèrcia rotacional, energia cinètica i principi de conservació de l'energia. **Donat:** * Massa del bloc (m) = 5 kg * Longitud de la corda (L) = 20 cm = 0,2 m (convertit a metres) * Velocitat angular final (ω_f) = 2 rad/s **Càlculs:** 1. **Moment d'inèrcia (I):** Com que el bloc gira al voltant d'un extrem de la corda, actua com una massa puntual a l'extrem d'una vareta. El moment d'inèrcia (I) d'aquest sistema és: I = mL^2 I = (5 kg) * (0,2 m)^2 I = 0,2 kgm^2 2. **Energia cinètica (KE):** Un cop el bloc comença a girar, guanya energia cinètica. L'energia cinètica d'un objecte en rotació és: KE = 1/2 * I * ω_f^2 KE = 1/2 * (0,2 kgm^2) * (2 rad/s)^2 KE = 0,4 J 3. **Treball fet (W):** Suposant que no hi ha pèrdua d'energia per fricció, el treball realitzat (W) per fer girar el bloc és igual a l'energia cinètica final guanyada. W = KE W = 0,4 J ## Mantenint la velocitat **No es requereix cap treball addicional per mantenir la velocitat angular de 2 rps un cop aconseguit.** Heus aquí per què, basant-nos en el principi de conservació de l'energia: * **Conservació de l'energia mecànica:** Aquest principi estableix que l'energia mecànica total (energia cinètica + energia potencial) en un sistema tancat es manté constant. * **Inicial i finalの状態 (joutai, estat):** Inicialment, el bloc està en repòs sobre una superfície horitzontal. Per tant, la seva energia cinètica és zero. Pot tenir una mica d'energia potencial a causa de la seva posició relativa a un punt de referència, però això no és rellevant aquí ja que la superfície és horitzontal. L'estat final té el bloc girant amb energia cinètica (0,4 J) però sense canvis en l'energia potencial. * **Sense pèrdua d'energia:** Com que la superfície no té fricció, no hi ha dissipació d'energia a causa de la fricció. Per tant, segons el principi de conservació de l'energia, una vegada que el bloc assoleixi una velocitat angular de 2 rps i la seva energia cinètica esdevingui 0,4 J, mantindrà aquesta velocitat indefinidament sense cap entrada de treball addicional sempre que el sistema romangui sense fricció i tancat (és a dir, , no hi actuen forces externes).

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