AP EAMCET · PHYSICS · Work Power Energy
A block of mass \(2 \mathrm{~kg}\) is initially at rest on a horizontal frictionless surface. A horizontal force \(\overrightarrow{\mathbf{F}}=\left(9-x^2\right) \hat{\mathbf{i}}\) newtons acts on it, when the block is at \(x=0\). The maximum kinetic energy of the block between \(x=0\) and \(x=3 \mathrm{~m}\) in joule is
- A \(24\)
- B \(20\)
- C \(18\)
- D \(15\)
Answer & Solution
Correct Answer
(C) \(18\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{KE}=\int F d x \\ & \qquad \begin{aligned} \mathrm{KE} & =\int_{x=0}^{x=3}\left(9-x^2\right) d x \\ & =9 \int_{x=0}^{x=3} d x-\int_{x=0}^{x=3} x^2 d x \\ & =9[x]_0^3-\left[\frac{x^3}{3}\right]_0^3=27-9=18 \mathrm{~J} .\end{aligned}\end{aligned}\)
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