Coefficient of kinetic friction between the two blocks

AI Thread Summary
The coefficient of kinetic friction between the two blocks is 0.30, with a frictionless table and pulleys. The masses involved are m1=2 kg, m2=3 kg, and m3=10 kg. Initial calculations yield an acceleration of -6.15 m/s², which differs from the book's answer of -5.7 m/s². Adjusting the equations leads to a recalculated acceleration of approximately 5.76 m/s². The discrepancy highlights the importance of correctly applying the frictional force in the equations of motion.
Squeezebox
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Homework Statement


The coefficient of kinetic friction between the two blocks is 0.30. The surface of the table and pulleys are frictionless.

m1=2 kg
m2=3 kg
m3=10 kg

Homework Equations


\SigmaF=ma
T2-m3g=m3a
T1-T2=m2a
friction-T1=m1a


The Attempt at a Solution


T2=m3g+m3a
T1=friction-m1a

friction-m1a-m3g-m3a=m2a

(friction-m3g)/(m1+m2+m3)=a
a=-6.15m/s2

The answer in my book say it is -5.7m/s2, though.
 

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Squeezebox said:

Homework Statement


The coefficient of kinetic friction between the two blocks is 0.30. The surface of the table and pulleys are frictionless.

m1=2 kg
m2=3 kg
m3=10 kg
The answer in my book say it is -5.7m/s2, though.
T2=m3g-m3a
T1=m1a+m1gμ
T2=T1+m1gμ+m2a
a=(m3g-m1gμ-m1gμ)/(m1+m2+m3)=5.7552
 

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