The block with mass m2=0.50kg is found to have a speed of 0.30m/s after it has dropped 0.80m. how large a (kinetic) friction force retards the motion of the block with mass m1=2.0kg?(adsbygoogle = window.adsbygoogle || []).push({});

(side note from teacher)

To solve this problem, you can use a string constraint. the total length of the string (L) is constant. However, as block m1 moves to the right, the portion of the string that is horizontal shortens by delta y and this length is distributed over the portion of the string taht is vertical according to the relationship; |delta x|=1/2|delta y|. By taking a derivative with respect to time twice, we have a1=2a2.

i think i did this completely wrong

i drew a free body diagram for each,for 1st free body diagram (mass1) and found m1 had Fn of 19.6

on the 2nd free body diagram (mass 2), i had Fg-Ft = ma

then, 4.9 N - 2Ft =0 , Ft = 2.45, then i said the Ft in the first free body diagram was also equal to 2.45, and therefore the kinetic friction force equals 2.45

its probably wrong, since i didnt use any info my teacher gave in the side note

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# Friction problem with pulleys

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