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The acceleration of system of two bodies...

The acceleration of system of two bodies over the wedge as shown in figure is `(m_(1) = 3 kg) (m_(2) = 5 kg)`

A

`3.39 m//s^(2)`

B

`4 m//s^(2)`

C

`1 m//s^(2)`

D

`2.5 m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`m_(2)g sin 60^(@) -T = m_(2)a`………(i)
`T- m_(1)g sin 30^(@) = m_(1) a`……..(ii)
Adding both equations:

`(m_(2) sin 60^(@) - m_(1) sin 30^(@))/(m_(1) + m_(2)) = (9.8 (5 xx sqrt(3)/2 -3 xx 1/2))/8`
`a=(9.8(2.5 xx 1.71 - 1.5))/8`
`a = 3.39 m//s^(2)`
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