Kinetic Friction of shoved box of books

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SUMMARY

The discussion focuses on calculating the time it takes for a box of books weighing 300 N to move 4.00 m when subjected to a 400 N force at a 35.2-degree angle below the horizontal, with a coefficient of kinetic friction of 0.570. The user derived the acceleration as -3.83 m/s² using the equations of motion and friction. The final calculation determined that the time required for the box to cover the distance is 1.44 seconds. The importance of drawing a free body diagram to visualize the forces acting on the box was emphasized for clarity in problem-solving.

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BBallman_08
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Homework Statement



A box of books weighing 300 N is shoved across the floor of an apartment by a force of 400 N exerted downward at an angle of 35.2 degrees below the horizontal. If the coefficient of kinetic friction between the box and the floor is .570, how long does it take the box to move 4.00 M, starting from rest.


Homework Equations



E Fx = 400 cos (35.2) - mK Fn = 300 (a)

E Fy = Fn - Fmg - 400 sin (35.2)



The Attempt at a Solution



400 cos (35.2) -(.570) (300) (9.8) + 400 Sin (30) = 300 A

A= -3.83 m/s2

4m = (1/2) (-3.83) (t^2)

t= 1.44s







This is what I have so far... I don't know if its correct?? Can someone help me out and check it? Thanks!
 
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BBallman_08 said:

Homework Statement



A box of books weighing 300 N is shoved across the floor of an apartment by a force of 400 N exerted downward at an angle of 35.2 degrees below the horizontal. If the coefficient of kinetic friction between the box and the floor is .570, how long does it take the box to move 4.00 M, starting from rest.


Homework Equations



E Fx = 400 cos (35.2) - mK Fn = 300 (a)

E Fy = Fn - Fmg - 400 sin (35.2)



The Attempt at a Solution



400 cos (35.2) -(.570) (300) (9.8) + 400 Sin (30) = 300 A

A= -3.83 m/s2

4m = (1/2) (-3.83) (t^2)

t= 1.44s







This is what I have so far... I don't know if its correct?? Can someone help me out and check it? Thanks!


400 cos (35.2) -(.570) (300) (9.8) + 400 Sin (30) = 300 A

This equation looks funny to me.. can you elaborate on how you got this?

[Edit]

I forgot to mention this, but on these type of problems, it is essential that you draw a free body diagram. Scan and attach one if you drew it.
 
Last edited by a moderator:
I substituted arranged the E Fy to solve for FN and then substituted that in for the FN in my Efx equation that I have listed up top.

Thanks for the help! I don't have a scanner to put my sketch on but I have the applied force coming in from the left side from above the block at an angle of 30 degrees below horizontal, the normal force going up from the box and the Fgrav headed down. The last force I ID'd was the F kf going to the left from the box.
 

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