Question

3. You have two textbooks hooked to a pulley system by light cord. Assume a light, frictionless pulley. The textbook on the table masses 2.50 kg. The hanging book masses 2.10 kg Α

I'm not sure about this problem, who can help me about this one.

Answer 1

$T=M_{A}a$ --(1)

--(2)

(As M_B is moving downwards M_B should be greater than tension)

adding equation (1) and (2) we get,

$a = \frac{M_{B}g}{M_{A}+M_{B}} = \frac{2.1*9.8}{2.1+2.5} = 4.474\ m/s^{2}$