Answer to Question #107743 in Classical Mechanics for Avalon

Question #107743
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?
1
Expert's answer
2020-04-06T08:55:48-0400

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}=\\\\\n\\space\\\\\n=a+\\frac{m_T}{m_S}(a+\\mu g)=6.14\\text{ m\/s}^2."


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