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A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 170 N at an angle of 25◦ above the horizontal. The box has a mass of 27.4 kg, and the coefficient of friction between box and floor is 0.293. The acceleration of gravity is 9.8 m/s 2 . 27.4 kg µ = 0.293 170 N 25◦ Find the acceleration of the box. Answer in units of m/s 2 .

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valeriagonzalez0213

Answer:

a = 3.52 m/s²

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Data

m= 27.4 kg : mas of the box

F= 170 N, at an angle of 25â—¦ above the horizontal :Force rope attached to the box

μk =  0.293 :Coefficient of friction between box and floor

g =   9.8 m/s² : acceleration due to gravity

We define the x-axis in the direction parallel to the movement of the  box and the y-axis in the direction perpendicular to it.

Forces acting on the box

W: Weight of the block : In vertical direction  ,downward

FN : Normal force : perpendicular to the floor  upward

f : Friction force: parallel to the floor  and opposite to the movement

F : force of the rope attached to the box , at an angle of 25â—¦ above the horizontal

Calculated of the W  ( weight of the box)

W= m*g

W=  27.4 kg* 9.8 m/s² = 268.52 N

x-y components  of the force of 170 N

Fx=170 N *cos 25° = 154.07 N

Fy=170 N *sin 25° =71.845 N

Calculated of the FN  ( Normal force)

We apply the formula (1)  

∑Fy = m*ay ay = 0  

FN + Fy - W = 0  

FN = W-  Fy

FN = 268.52 N -   71.845 N

FN =196. 675 N

Calculated of the f  (friction force)

f = μk*FN

f = 0.293*196. 675

f = 57.626 N

We apply the formula (1) to calculated acceleration of the box:

∑Fx = m*ax  ,  ax= a  : acceleration of the box

Fx-f = m*a

154.07-57.626 = (27.4)*a

96.45 =  (27.4)*a

a = (96.45)/ (27.4)

a = 3.52 m/s²

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