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- Homework Statement
- [Young & Freedman - University Physics 13th ed, exercise 6.87]

Consider the system shown in Fig. P6.86. The rope and pulley have negligible mass, and the pulley is frictionless. Initially the 6.00-kg block is moving downward and the 8.00-kg block is moving to the right, both with a speed of 0.900 m/s. The blocks come to rest after moving 2.00 m. Use the work–energy theorem to calculate the coefficient of kinetic friction between the 8.00-kg block and the tabletop.

- Relevant Equations
- Work-energy theorem

Kinetic friction formula

My final answer is different from the official one in the back of the book, and I can't figure out what I did wrong. This is my attempt:

Let

**block 1**be the vertically moving block and let

**block 2**be the horizontally moving one.

Also, let ##m_1 = 6.00 ~\rm{kg}##, ##m_2 = 8.00 ~\rm{kg}##, ##v_0 = 0.900 ~\rm{m/s}##, and ##d = 2.00 ~\rm{m}##. The target variable is ##\mu_k##.

According to the work-energy theorem, the total work done on block 2 is ##W_{tot}= \Delta K = K_f - K_i = 0 -K_i = -\frac{1}{2}m_2v_0^2##.

Alternatively, since the forces acting on block 2 are constant, we also have ##W_{tot} = \Sigma F_x d = (T - f_k)d = (m_1g - \mu_km_2g)d## (if the direction of displacement is the positive direction).

So then we have

$$-\frac{1}{2}m_2v_0^2 = (m_1g - \mu_km_2g)d$$

Rearranging to express ##\mu_k##, we obtain

$$\mu_k = \frac{m_1}{m_2} + \frac{v_0^2}{2dg} = \frac{6.00 ~\rm{kg}}{8.00 ~\rm{kg}} + \frac{(0.900 ~\rm{m/s})^2}{2(2.00 ~\rm{m})(9.80 ~\rm{m/s^2})} = 0.771 $$

However, the book gives ##0.786##. Where did I mess up?