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Whitepunk [10]
4 years ago
13

Una persona y su perro se encuentran inicialmente en reposo y paradas arriba de una caja que está apoyada sobre un piso rugoso.

La masa total de la caja, la persona y la mascota es de 80 kg. El coeficiente de rozamiento cinético entre la caja y el piso vale 0,3. Cuando el perrito salta hacia el piso con una velocidad horizontal de 6,0 m/s, la caja y la persona se desplazan 1,0 m en sentido contrario. ¿Cuál es la masa de la mascota?
Physics
1 answer:
alexandr1967 [171]4 years ago
5 0

Answer:

La masa de la mascota es de 11,24 kg.

Explanation:

Dado que la masa total de = 80 kg

Tenemos

Fuerza de fricción = 80 kg × 9.81 × 0.3 = 235.44 N

La mascota salta horizontalmente con una velocidad, v, de 6.0 m / s para dar un movimiento de 1 m a la caja de masa y a la persona

Trabajo realizado por el salto de la mascota = (80 - m) × 9.81 × 0.3

La energía del movimiento del perro = 1/2 × m × v² = Trabajo realizado al saltar de la mascota

1/2 × m × 6² = (80 - m) × 9.81 × 0.3

20943 · m = 235440

m = 235440/20943 = 11,24 kg

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Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

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