TS EAMCET · Physics · Work Power Energy
A body of mass \(2.4 \mathrm{~kg}\) is subjected to a force which varies with distance as shown in figure. The body starts from rest at \(x=0\). Its velocity at \(x=9 \mathrm{~m}\) is

- A \(5 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
- B \(20 \sqrt{3} \mathrm{~m} / \mathrm{s}\)
- C \(10 \mathrm{~m} / \mathrm{s}\)
- D \(40 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(10 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Work done in a moving body is get converted into kinetic energy. \( \begin{aligned} & \therefore & \int F d x & =\frac{1}{2} m\left(v^2-0^2\right) \\ & \Rightarrow & \frac{1}{2}(9+3)(20) & =\frac{1}{2}(2.4) v^2 \\ & \Rightarrow & V & =10 \mathrm{~m} / \mathrm{s} \end{aligned} \)
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