Answer to Question #333202 in Mechanics | Relativity for Sujit

Question

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on the top of the slab. The coefficient of static

friction is 0.6 while the coefficient of kinetic friction is 0.4. 10 kg block is acted upon by a horizontal force

F. What is the resulting acceleration of the slab if (i) F = 40 N

Answer 1

$a=\frac {F+\mu_1m_1g+\mu_2m_2g}{m_1+m_2}=3.5~\frac{m}{s^2}.$