Answer:
(b)
\[{{F}_{x}}=3N,\text{ }{{F}_{y}}4N\]
\[{{\upsilon
}_{ax}}\,=6\,m{{s}^{-1}},\,\,{{\upsilon }_{ay}}=-12\,m{{s}^{-1}}\]
Now,
\[{{a}_{x}}\,=\frac{{{F}_{x}}}{m}=\,\frac{-3}{5}\,m{{s}^{-2}}\]
Using
\[{{\upsilon }_{x}}\,={{\upsilon }_{ax}}+\,at=6\,-\frac{3}{5}\,t\]
Since,
\[{{\nu }_{x}}\,=0\,\,\therefore \,\frac{3}{5}\,t\,\,=6\] or \[t=10s\,\,\]
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