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What is the acceleration of the 3.0kgblock in FIGURECP7.55across the frictionless table?

Hint: Think carefully about the acceleration constraint.

Short Answer

Expert verified

The acceleration of the 3.0kgblock is 2.8m/s2

Step by step solution

01

Given information

mass of blockm1=3.0kg

mass of blockm2=1.0kg

All surfaces are frictionless.

02

Explanation

Consider the Forces of two masses given in thefigureCP7.55

For block massm1

T1=m1a1

For block massm2

m2g-T2=m2a2

Considering the pulley is massless, T1=T2

Let a2=aT=2m1a

a1=2aacceleration constraint

On substituting the above equations on block massm2equation, we get

m2g-2m1a=m2a(2m1+m2)a=m2ga=m2g(2m1+m2)

Therefore, the acceleration for m1is given by,

localid="1649865147826" a1=2m2g(2m1+m2)=2(1.0)(9.81)(2(3.0)+1.0)=2.8m/s2

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