A) 10 N
B) 20 N
C) 32 N
D) 40 N
Correct Answer: C
Solution :
Here: \[{{m}_{1}}=10kg,{{m}_{2}}=6kg,{{m}_{3}}=4kg,\] \[F=40N\] Since, the table is frictionless i.e., it is smooth therefore, force on the blocks is given by \[F=({{m}_{1}}+{{m}_{2}}+{{m}_{3}})a\] \[\Rightarrow \] \[a=\frac{F}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] \[=\frac{40}{10+6+4}=\frac{40}{20}\] \[=2\text{ }m/{{s}^{2}}\] Now the tension between 10 kg and 6 kg masses is given by \[{{T}_{2}}=({{m}_{1}}+{{m}_{2}})a\] \[=(10+6)2=16\times 2\] \[=32N\]You need to login to perform this action.
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