A) 98.596 N
B) 985.96 N
C) 100.2 N
D) 76.23 N
Correct Answer: A
Solution :
Maximum acceleration \[{{A}_{\max }}\]\[=a{{\omega }^{2}}=\frac{a\times 4{{\pi }^{2}}}{{{T}^{2}}}\]\[=\frac{1\times 4\times {{(3.14)}^{2}}}{0.2\times 0.2}\] \[{{F}_{\max }}=m\times {{A}_{\max }}=\frac{0.1\times 4\times {{(3.14)}^{2}}}{0.2\times 0.2}=98.596\ N\]
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