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Two blocks of masses m(1)=1 kg and m(2)=...

Two blocks of masses `m_(1)=1 kg and m_(2)=2 kg` are connected by a string and side down a plane inclined at an angle `theta=45^(@)` with the horizontal. The coefficient of sliding friction between `m_(1)` and plane is `mu_(1)=0.4,` and that between `m_(2)` and plane is `mu_(2)=0.2.` Calculate the common acceleration of the two blocks and the tension in the string.

Text Solution

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As `mu_(2)ltmu_(1),` block `m_(2)` has greater acceleration than `m_(1)` if we separately consider motion of blocks. But they are connected so they move together as a system with common acceleration. So acceleration of the blocks :
`a=((m_(1)+m_(2))gsintheta-mu_(1)gcostheta-mu_(2)m_(2)gcostheta)/(m_(1)+m_(2))`
`=((1+2)(10)((1)/(sqrt2))-0.4xx1xx10xx(1)/(sqrt2)-0.2xx2xx10xx(1)/(sqrt2))/(1+2)=(22)/(3sqrt2)ms^(-2)`
For block `m_(2):m_(2)gsintheta-mu_(2)m_(2)gcostheta-T=m_(2)aimpliesT=m_(2)gsintheta-mu_(2)m_(2)gcostheta-m_(2)a`
`=2xx10xx1/sqrt2-0.2xx2xx10xx1/sqrt2-2xx(22)/(3sqrt2)=(2)/(3sqrt2)N`
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