Question

A fixed Pulley with a smooth grove has a light string passing over it with a 4 kg attached on one side and a 3kg on the other side. Another 3 Kg is hung from the other 3 kg as shown in the figure with another light string. If the system is released from rest, find the common acceleration? (g = 10 m/s²)

Answer 1

Hii dear,

# Answer-
a = 2 m/s^2

## Explaination-
# Given-
m1 = 4 kg
m2 = 3+3 = 6 kg
g = 10 m/s^2

# Solution-
Suppose, two masses be m1 (4kg) and m2(3+3 kg) such that m2>m1.
Also T = tension in string
a = acceleration of m1 (upwards) & m2 (downwards).

Let's get some equations,
T - m1g = m1a ...(1)
m2g - T = m2a ...(2)

Adding eqns & solving them
a = (m2-m1)g/(m1+m2)
a = (6-4)×10/(6+4)
a = 2 m/s^2

Hence, When realesed, acceleration in the string will be 2 m/s^2.

Hope this is useful...

Answer 2

Answer:

2ms^-2

Explanation:

Given figure looks like an Atwood's machine

Therefore, m1 =4kgs

m2=(3+3)kgs =6 kgs.

Common acceleration, a=(m2-m1/m1+m2)g

= a = (6-4/6+4)10 = 2 ms ^-2

Hope it helps you