JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(.............\,j\).
- A \(131\)
- B \(134\)
- C \(133\)
- D \(132\)
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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