A) \[\text{24 }m/{{s}^{2}}\]
B) \[\text{3}\sqrt{71}\text{ }m/{{s}^{2}}\]
C) \[\text{30 }m/{{s}^{2}}\]
D) \[\text{25 }m/{{s}^{2}}\]
Correct Answer: D
Solution :
[d] When all are pulling \[{{\vec{F}}_{net}}=100\times 3\hat{i}\text{ }.....\text{(i)}\] When A stops \[{{\vec{F}}_{net}}-{{\vec{F}}_{A}}=100\times 1\left( -\hat{i} \right)....\left( ii \right)\] From these three we get \[{{\vec{F}}_{A}}+{{\vec{F}}_{B}}=100\left( 7\hat{i}-24\hat{j} \right)\] \[\therefore \text{ Required }ac{{c}^{n}}=\sqrt{{{7}^{2}}+{{\left( -24 \right)}^{2}}}=25\text{ m/}{{\text{s}}^{2}}\]You need to login to perform this action.
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