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Springs and Friction |
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| Dec6-06, 01:48 PM | #1 |
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Springs and Friction
1. The problem statement, all variables and given/known data
A 30.0 kg block is resting on a flat horizontal table. On top of this block is resting a 15.0 kg block, to which a horizontal spring is attached. The spring constant of the spring is 325 N/m. The coefficient of kinetic friction between the lower block and the table is 0.600, and the coefficient of static friction between the two blocks is 0.900. A horizontal force T is applied to the lower block toward the spring. This force is increasing in such a way to keep the blocks moving at a constant speed. At the point where the upper block begins to slip on the lower block, determine (a) the amount by which the spring is compressed and (b) the magnitude of the Force T 2. Relevant equations F=kx Ff= uma Fw=mg 3. The attempt at a solution we did a problem like this with out the spring and i did it like this Fk1=u1mg Fk2=u2u1mg Fk1+Fk2=FT u1mg+u1u2mg=FT .9(15)(9.8)+.6(.9)(15+30)(9.8)=FT 370.44=Ft taht migth be right i dont know if that is or how you ahd a spring putting a force on the top block please help me get started |
| Dec6-06, 02:05 PM | #2 |
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Start by identifying all the forces acting on the smaller block.
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| Dec6-06, 05:48 PM | #3 |
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goign to set right positive, oppsite of force T
Force on the spring=kx Force of weight= mg Force of kinetic energy= uma i think thats all i know |
| Dec6-06, 06:31 PM | #4 |
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Springs and Friction |
| Dec6-06, 06:42 PM | #5 |
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there is no acceleration so the net forces are equal to 0
the other forces would be the force is is being pushed toward the spring and maybe something to do with the friction force |
| Dec6-06, 08:12 PM | #6 |
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| Dec6-06, 09:13 PM | #7 |
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FT=Force of the Spring |
| Dec7-06, 11:56 AM | #8 |
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