A) 0.1 m/s
B) 1.5 m/s
C) 5 m/s
D) 60 m/s
Correct Answer: A
Solution :
From Newton's 2nd law \[F=ma\] ...(i) From equation of motion substituting \[{{v}^{2}}={{u}^{2}}+2as\] \[{{v}^{2}}=0+2as\] \[a=\frac{{{v}^{2}}}{2s}\] ?.(ii) From Eqs. (i) and (ii) \[F=\frac{m{{v}^{2}}}{2s}\] ?..(iii) Given, \[F=5\times {{10}^{4}}N\] \[m=3\times {{10}^{7}}kg\] \[s=3m\] \[\therefore \] \[5\times {{10}^{4}}=3\times {{10}^{7}}\times \frac{{{v}^{2}}}{2\times 3}\] \[{{v}^{2}}=\frac{5\times {{10}^{4}}\times 2}{{{10}^{7}}}\] or \[{{v}^{2}}={{10}^{-2}}m/s\] So, \[v={{10}^{-1}}m/s=0.1m/s\]
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