Laws Of Motion Question 260

Question: A body of mass 8kg is moved by a force $ F=3xN, $ where $ x $ is the distance covered. Initial position is $ x=2m $ and the final position is $ x=10 $ m. The initial speed is $ 0.0m/s. $ The final speed is[Orissa JEE 2002]

Options:

A) 6 m/s

B) 12 m/s

C) 18 m/s

D) 14 m/s

Show Answer

Answer:

Correct Answer: A

Solution:

Increment in kinetic energy = work done therefore

$ \frac{1}{2}m(v^{2}-u^{2})=\int _{x _{1}}^{x _{2}}{F.dx}=\int _{2}^{10}{(3x)\ dx} $

therefore $ \frac{1}{2}mv^{2}=\frac{3}{2}[x^{2}] _{2}^{10}=\frac{3}{2}[100-4] $

therefore $ \frac{1}{2}\times 8\times v^{2}=\frac{3}{2}\times 96 $

therefore $ v=6m/s $



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