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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2023
  • Medium
  • Numerical

A force \(F=\left(5+3 y^2\right)\) acts on a particle in the \(y\) direction, where \(F\) is newton and \(y\) is in meter. The work done by the force during a displacement from \(y =2\,m\) to \(y =5\,m\) is \(.............\,j\).

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Answer & Solution

Correct Answer

(D) \(132\)

Step-by-step Solution

Detailed explanation

\(F =5+3 y ^2\)

\(W =\int \limits_2^5\left(5+3 y ^2\right) d y\)

\(=\left[5 y +\frac{3 y ^3}{3}\right]_2^5\)

\(=132\,J\)

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