question_answer

A load of mass M = 1000 kg is supported in a vertical plane by a string and pulleys. If the free end A of the string is pulled vertically downward with an acceleration of \[4\text{ }cm/{{s}^{2}},\] the tension in the string is

A)  1000 N                      

B)  3000 N          

C)  4000 N                

D)       5000 N

Correct Answer: D

Solution :

Invoking work energy theorem on the system encapsulated by the dotted boundary: \[T{{a}_{A}}+2T{{a}_{B}}=0\] \[\Rightarrow \]\[{{a}_{A}}=-2{{a}_{B}}\] or,\[{{a}_{B}}=\frac{-{{a}_{A}}}{2}=\frac{-4}{2}=-2\frac{cm}{{{s}^{2}}}=\frac{-2}{100}\,m/{{s}^{2}}\] F. B. D of block B: \[1000\,g-2T=1000{{a}_{B}}\] \[2T=1000\,g-1000{{a}_{B}}\] \[\Rightarrow \]   \[T=500g-500\,{{a}_{B}}\] \[=500\times 10-500\left( \frac{-2}{100} \right)\] \[=5000+10=5010=5000\,N\] Hence, the correction option is [d].