A) 20 N
B) 40 N
C) 10 N
D) 32 N
E) 16 N
Correct Answer: D
Solution :
Common acceleration \[a=\frac{F}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] \[a=\frac{40}{10+6+4}=2\,m/{{s}^{2}}\] Equation of motion of\[{{m}_{3}}\]is \[{{T}_{3}}-{{T}_{2}}={{m}_{3}}a\] \[40-{{T}_{2}}=4\times 2\] \[\Rightarrow \] \[{{T}_{2}}=32N\]You need to login to perform this action.
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