Answer:
18 N
Step-by-step 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.