Question

5. Consider the system shown in the figure. Block A has a weight 𝑤𝐴 and block B has a weight 𝑤𝐵. Once block B is set into downward motion, it descends at a constant speed. a) Calculate the coefficient of kinetic friction between block A and the tabletop. b) A cat, also of weight 𝑤𝐴, falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration (magnitude and direction)?

Answer 1

(a) Deceleration with constant speed means equilibrium of forces:

$$B:W_B=T,\\\space\\ A:T=\mu N=\mu W_A.\\\space\\ \mu=\frac T{W_A}=\frac{W_B}{W_A}.$$

(b) Write equations by Newton's second law for the scenario with cat:

$$B:T-W_B=-\frac{W_B}ga,\\\space\\ A: T-2\mu W_A=\frac{W_A}ga.\\\space\\ T=W_B-\frac{W_B}ga,\\\space\\ W_B-\frac{W_B}ga-2\frac{W_B}{W_A}W_A=\frac{W_A}ga,\\\space\\ a\bigg(\frac{W_A}g+\frac{W_B}g\bigg)=-W_B,\\\space\\ a=-g\frac{W_B}{W_A+W_B}.$$