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Two masses as shown are suspended from a massless pulley. Calculate the acceleration of the system when masses are left free :
A
$\dfrac{2g}{3}$
B
$\dfrac{g}{3}$
C
$\dfrac{g}{9}$
D
$\dfrac{g}{7}$
Detailed Solution
The heavier 10 kg mass moves down and the 5 kg mass moves up with the same acceleration a. Let T be the tension in the string.
[add image: free body diagrams, T up and 10g down on the 10 kg mass; T up and 5g down on the 5 kg mass]
For the 10 kg mass: $10g - T = 10a$
For the 5 kg mass: $T - 5g = 5a$
Adding: $5g = 15a$
$a = \dfrac{g}{3}$
(In general $a = \dfrac{(m_1 - m_2)g}{m_1 + m_2}$.)
[add image: free body diagrams, T up and 10g down on the 10 kg mass; T up and 5g down on the 5 kg mass]
For the 10 kg mass: $10g - T = 10a$
For the 5 kg mass: $T - 5g = 5a$
Adding: $5g = 15a$
$a = \dfrac{g}{3}$
(In general $a = \dfrac{(m_1 - m_2)g}{m_1 + m_2}$.)