Answer to Question #107743 in Classical Mechanics for Avalon

Question

Two figure skaters, Tessa (55kg) and Scott (83kg) are facing one another during their program. Scott pushes Tessa, accelerating her at 3.5m/s^2. If the coefficient of friction on the ice is 0.05, What is Scott's acceleration?

Answer 1

Since they both accelerate at 3.5 m/s/s, the net force is

$$F_{net}=(m_S+m_T)a.$$

According to Newton's second law, this resultant force equals Scott's thrust minus Tessa'a force of friction:

$$F_{net}=F_S-f_T=F_S-\mu m_Tg.$$

Hence:

$$a_S=\frac{F_S}{m_S}=\frac{(m_S+m_T)a+\mu m_Tg}{m_S}=\\ \space\\ =a+\frac{m_T}{m_S}(a+\mu g)=6.14\text{ m/s}^2.$$