two bodies of mass 3 kg and 4kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is(take g= 9.8 m/s2)
a) 4.9 m/s2
b)2.45 m/s2
c)1.4 m/s2
d)9.5 m/s2
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hey dude..
ur answer is here....!!!!!!!
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
ur answer is here....!!!!!!!
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
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