How long does it take to move a box of books across the floor?

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SUMMARY

The discussion centers on calculating the time required to move a box of books weighing 329 N across a floor, using a force of 484 N applied at a 32° angle. The coefficient of kinetic friction (μk) is 0.57, and the acceleration due to gravity is 9.81 m/s². The correct approach involves calculating the normal force and kinetic friction, leading to an acceleration of approximately 2.03 m/s². The final calculation yields a time of approximately 1.90 seconds to move the box 3.66 m, starting from rest.

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Homework Statement


A box of books weighing 329 N is shoved across the floor by a force of 484 N exerted downward at an angle of 32◦ below the hori- zontal.
The acceleration of gravity is 9.81 m/s2 .
If μk between the box and the floor is 0.57, how long does it take to move the box 3.66 m, starting from rest?
Answer in units of s.

Homework Equations


Force=Mass(Acceleration)
X=1/2aT^2

The Attempt at a Solution


(Sigma)Fx=F(friction)+F(Applied)=ma
(Sigma)Fy=w+F(Normal)+F(Applied)=0

F(Normal)=329N + 484N(Sin32) = 585.481 ---->
F(Kinetic) = Mu(Kinetic)*F(Normal) ---> Fk = 0.57(585.481) = 333.724

-F(Kinetic)+F(Applied)(Cos32)
---------------------------------- = a -----> m=F(gravity)/(9.81)
m

-333.724 + 484(Cos32)
-------------------------- = 2.02893
329x=1/2at^2
3.66=1/2(2.02893)t^2
t=1.89942

I'm pretty sure I've done this correct, but the answer keeps coming out incorrect. Somebody please check my work.
 
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It looks like you used 329 as the mass instead of 329/9.81.
And something else went wrong in the same step to get a = 2.02.
Anyway, I get acceleration about 10% higher.
 

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