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alexandr402 [8]
3 years ago
14

Se lanza un cuerpo de 25 kg sobre un plano horizontal, con una velocidad de 20 m/s. Si el coeficiente de rozamiento es de 0,4 ca

lcula el valor de la fuerza de rozamiento y el espacio que recorrerá el cuerpo hasta pararse
Physics
1 answer:
Alex Ar [27]3 years ago
8 0

Answer:

1. 98 N

2. 51.0 m

Explanation:

masa 25 kg

velocidad 20 m/s

coeficiente de rozamiento .4

<u>Fuerza de rozamiento:</u>

mu * N, where mu = coeficiente de rozamiento, N = fuerza normal

Calcula N:

F = ma = 25 * -9.8 = -245 N

F rozamiento = mu * N = .4 * -245 = -98 N

<u>Espacio que recorrerá:</u>

F rozamiento = ma

-98 = 25a

a = -3.92 de-aceleración de la fuerza

vf = 0 m/s

vf = at + vi

0 = -3.92t + 20

t = 5.10 seg

x = 1/2 at^2 + vi*t

  = 1/2 * -3.92 * 5.10^2 + 20*5.10

  = 51.0 m

 

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1. Largest force: C;  smallest force: B; 2. ratio = 9:1

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\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}}  - \dfrac{k}{(2d)^{2}}  +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

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\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}}  + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

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\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}}  - \dfrac{k}{(2d)^{2}}  - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

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2. Ratio of largest force to smallest

\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}

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