A) 6 m/s
B) 12 m/s
C) 18 m/s
D) 14 m/s
Correct Answer: A
Solution :
Increment in kinetic energy = work done Þ \[\frac{1}{2}m({{v}^{2}}-{{u}^{2}})=\int_{{{x}_{1}}}^{{{x}_{2}}}{F.dx}=\int_{2}^{10}{(3x)\ dx}\] Þ \[\frac{1}{2}m{{v}^{2}}=\frac{3}{2}[{{x}^{2}}]_{2}^{10}=\frac{3}{2}[100-4]\] Þ \[\frac{1}{2}\times 8\times {{v}^{2}}=\frac{3}{2}\times 96\] Þ \[v=6m/s\]
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