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Ghella [55]
3 years ago
14

Un bloque de 2.5kg de masa es empujado 2.2m a lo largo de una mesa horizontal sin fricción por una fuerza constante de 16.0 N di

rigida a 25° debajo de la horizontal. Encuentre el trabajo efectuado por: (a) la fuerza aplicada, (b) la fuerza normal ejercida por la mesa, (c) la fuerza de la gravedad, y (d) la fuerza neta sobre el bloque

Physics
1 answer:
Tanzania [10]3 years ago
8 0

Answer:

Explanation:

(a) The applied force has two components Fx and Fy. The Fx component is the only one that does work

W_{x}=F_{x}x=(16N)cos(25)(2.2m)=31.9J

(b) There in no net force in the vertical component

F_{N}-F_{y}-F_{g}=0\\F_{N}=F_{y}+F_{g}=(16N)sin(25)+(2.5kg)(9.8\frac{m}{s^{2}})=31.26N

(c)

F_{g}=Mg=(2.5kg)(9.8\frac{m}{s^{2}})=24.5N

(d)

F_{T}=F_{x}=(16N)cos(25)=14.5N

I attached an scheme of the force diagram

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