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 $
BETA
