What's the downward acceleration?

Question

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

Answer 1

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.

 

Answer 2

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