Posted by ericka on Monday, May 7, 2012 at 10:42pm.
A 1.5kg block rests on top of a 7.5kg block. The cord and pulley have negligible mass, and there is no significant friction anywhere.
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

physics  Elena, Tuesday, May 8, 2012 at 11:36am
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 = TF,
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.
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