A constant force F = m 2 g/2 is applied on the block of mass m 1 as shown in fig. The string and the pulley are light and the surface of the table is smooth. The acceleration of m 1 is –
A
towards right
B
towards left
C
towards right
D
towards left
Text Solution
Verified by ExpertsThe correct answer is:
A
m 1 : T – F = m 1 a
F = m 2 g/2
∴ T = 
+ m 1 a .... (i)
m 2 : m 2 g – T = m 2 a ⇒ T = m 2 g – m 2 a .... (ii)
(i) = (ii)
∴ 
+ m 1 a = m 2 g – m 2 a ⇒ a = 
short trick
a = 
= 
= 
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