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Papessa [141]
3 years ago
12

A college student likes to take her books to class by placing them in a box and pulling the box around behind her. She pulls on

the box with a force of 90.0 N at an angle of 30.0⁰ relative to the horizontal. The box of books has a mass of 20.0 kg and the coefficient of kinetic friction between the bottom of the box and the sidewalk is 0.500. What is the net force acting horizontally and what is the acceleration of the box as she pulls it around?
Physics
1 answer:
just olya [345]3 years ago
8 0

Answer:

It would result an a negatice answer.

Explanation:

The accelarion should be pulled as a kite not a box :) columbus said that musical stuff no just no

Hope this helps!!

       - Katty queen

             

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Answer:

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Explanation:

Look at the attached graphic

we take the forces acting on the x-y plane and applied at the origin of coordinates

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x-y components FY

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FYy= -  1530*sin(45)° = - 1081.87 P

Calculation of the components of net force (Fn)

Fnx= FX + FYx

Fnx= -1260 P -1081.87 P

Fnx= -2341.87 P

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Fny= -1081.87 P

Calculation of the components of equilibrant force (Fe)

the x-y components of the  equilibrant force are equal in magnitude but in the opposite direction to the net force components:

Fnx= -2341.87 P, then, Fex= +2341.87 P

Fny=  -1081.87 P P, then, Fex= +1081.87 P

Magnitude of the equilibrant (Fe)

F_{n} = \sqrt{(F_{nx})^{2} +(F_{ny})^{2}  }

F_{e} =\sqrt{(2341.87)^{2}+(1081.87)^{2}  }

Fe= 2579.68 P

Calculation of the direction of  equilibrant force (α)

\alpha =tan^{-1} (\frac{F_{ny} }{F_{nx} } )

\alpha =tan^{-1} (\frac{1081.87 }{2341.87} )

α= 24.8°

Look at the attached graphic

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