A 1.5-kg block rests on top of a 7.5kg block. The cord and pulley have negligible mass, and there is no significant friction anywhere.

Question

Part a) What force F must be applied to the bottom block so the top block accelerates to the right at 2.9 m/s^2 ?

Part b) What is the tension in the connecting cord

Answer 1

m1 =1.5 kg, m2 = 7.5 kg, a=2.9 m/s²
(a) The horizontal projections of the equations of motion for each block are
m1•a = T,
m2•a = T-F,
F = (m1+m2) •a = (1.5+7.5) •2.9 = 26.1 N,
(b) T= m2•a - F= 7.5•2.9 – 26.1 = 4.35 N.