Hi if the coefficient of kinetic friction between the 5.0 kg mass and the table is 0.305, what is the tension in the string?

18 N

Explanation:

Draw a free body diagram for each block.

There are four forces acting on block I:

Weight force Mg pulling down

Normal force N pushing up

Tension force T pulling right

Friction force Nμ

There are two forces acting on block II:

Weight force mg pulling down

Tension force T pulling up

Sum of forces on block I in the +y direction:

∑F = ma

N − Mg = 0

N = Mg

Sum of forces on block I in the +x direction:

∑F = ma

T − Nμ = Ma

T − Mgμ = Ma

Sum of forces on block II in the -y direction:

∑F = ma

mg − T = ma

Solve for a in the first equation, then substitute into the second.

a = (T − Mgμ) / M

mg − T = m (T − Mgμ) / M

mMg − MT = mT − mMgμ

mMg + mMgμ = mT + MT

mMg (1 + μ) = (m + M) T

T = mMg (1 + μ) / (m + M)

T = (2) (5) (9.8) (1 + 0.305) / (2 + 5)

T = 18.27

Rounding to two significant figures, the tension is 18 N.

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