The coefficient of kinetic friction between the 2.0kgblock in FIGURE P7.38 and the table is 0.30. What is the acceleration of the 2.0kg block?

Short Answer

Expert verified

Acceleration of the block is6.91m/s2.

Step by step solution

01

Given information

Coefficient of friction between 2kgblock and table, μ=0.3

Given the figure of three blocks

02

Explanation

Consider the net force applied by 1kgblock

localid="1649647806258" F1=1kg×9.81m/s2=9.81Nrighttoleft

The friction force applied to the block 2

localid="1649647844745" F2=2kg×0.3×9.81m/s2=5.8Nrighttoleft

The force applied by the block 3

localid="1649647866757" F3=3kg×9.81m/s2=29.43Nlefttoright

Net force o block 2

localid="1649647898415" F=29.43N-5.8N-9.81N=13.82N

The acceleration of the block is calculated as

localid="1649647916525" F=ma13.82N=2kg×aa=6.91m/s2

Acceleration of the block is6.91m/s2.

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