Question

Two bodies of mass 3 kg and 4 kg are suspended at the ends of
massless string passing over a frictionless pulley. The
acceleration of the system is (g = 9.8 m/s2)
[MP PET 1994; CBSE PMT 2001]
(a) 4.9 m/s2
(b) 2.45 m / s?
(c) 1.4 m/s2
(d) 9.5 m/s2​

Answer 1

Answer:

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When the bodies are released, the heavier one moves downward and the lighter one moves upward.

Net force in the direction of motion of

m1 is F 1 = m1 g − T which is equal to m1 a

as per Newton's second law.

⇒ m1 g − T = m1 a ____________(i )

Net force in the direction of motion of

m2 is F 2 = T − m2 g which is equal to m2 a

as per Newton's second law.

⇒ T − m2 g = m2 a ___________(ii )

Solving (i ) and (ii ) for acceleration,

a → a = m1 − m2

m1 + m2 g

In our case,

m1 = 4, m2 = 3

→ a = 4−3

4+3× 9.8

= 1.4 m / s 2