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limpexowns

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## Homework Statement

(1) Firstly, the block as it rest at incline of 10 degrees and mass is added on the other side of pulley to overcome static friction (0.0063kg). The block weighs 0.8kg. Calculate the tension.

(2) The block is at same angle and mass but this time the block was lightly pushed from back to make it move. As the block began to move, mass was added on the other side of the pulley to just keep moving it forward. The mass was (0.00480kg). Calculate Tension.

## Homework Equations

T=mg (to calculate tension for static friction)

T=mg+Fs

Fs=T-Fgsin@

Fk=T-Fgsin@

## The Attempt at a Solution

For (1), I used T=mg to find tension which comes out to be 0.618N. I used that value to find Static Friction and I got 0.745N

For (2), I tried using T=mg to find Tension and I got 0.471N but when I used that value to find Kinetic friction, I'm getting 0.892N. This value didn't make sense because the Kinetic friction is always less than static friction but here it was higher.

So, I tried using T=mg+Fs for tension in Kinetic friction. I assumed that Fs was force applied because the cart was slightly pushed just o make it move. By using that, I got T=1.22N. I used that value to find Kinetic friction and I'm getting 0.143N. This value makes sense to me because the kinetic friction is lower than static.

The questions are very similar to this lab: http://ems.calumet.purdue.edu/chemphys/ncrelich/PortableDocuments/PHYS152/Summer2009/LabInclinedPlane.pdf

So, can you please let me know which method is right in order to calculate Tension for Kinetic part. Thanks a lot.

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