A 2.0 kg block has a rope attached to the block on a table and is pulled with a force of 8.0 N. The block accelerated at 2.5m/s^2. The coefficient of kinetic friction between the block and the table is . (Round to the nearest hundred.)

0.15

Explanation:

Assuming the rope is horizontal, sum the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum the forces in the x direction:

∑F = ma

F − Nμ = ma

Substitute:

F − mgμ = ma

mgμ = F − ma

μ = (F − ma) / (mg)

Plug in values:

μ = (8.0 N − 2.0 kg × 2.5 m/s²) / (2.0 kg × 9.8 m/s²)

μ = 0.15

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