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What's the downward acceleration?

Two masses are connected by a light cord which is looped over a light frictionless pulley. If one mass is 3.0 kg and the second mass is 5.0 kg, what's the downward acceleration of the heavier mass? Assume air resistance is negligible.
 
Answer: 2.5 m/s^2
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2 Answers

A quick way: Treat two masses as a whole system. The net downward force is (m1-m2)g, where m1 = 5 kg, and m2 = 3 kg.
 
Apply Newton's second law to the whole system:
(m1-m2)g = (m1+m2)a
 
Solve for a,
a = (m1-m2)g/(m1+m2) = (5-3)g/(5+3) = 2.45 m/sec2 <==Answer
Apply Newton's 2nd law to each mass:
 
For the 3.0-kg mass, which moves up,
 
3a=T-3g, where T is the tension in the cord.
 
For the 5.0-kg mass, which moves down,
 
-5a=T-5g.
 
Eliminate T and get
 
-5a=(3a+3g)-5g=3a-2g
2g=8a
a=g/4=9.8/4=2.45 m/s².