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Two masses of 5 kg and 3 kg are suspended with help of mass less inextensible string as shown. Calculate T_1 and T_2 when system is going upwards with acceleration m//s^2. ("Use g"= 9.8m//s^2).

Answer»

SOLUTION :According Newton’s second LAW of motion
(i) `T_(1)-(m_(1)+m_(2))g=(m_(1)+m_(2))a`
`T_(1)=(m_(1)+m_(2))(a+g)`
=(5+3) (2+9.8)
`T_(1)=94.4N`
(ii) `T_(2)-m_(2)g=m_(2)a`
`T_(2)=m_(2)(a+g)`
`T_(2_)=3(2+9.8)`
`T_(2)=35.4N`


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