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

A force \(\overrightarrow{ F }=(2+3 x) \hat{ i }\) acts on a particle in the \(x\) direction where \(F\) is in newton and \(x\) is in meter. The work done by this force during a displacement from \(x =0\) to \(x =4\,m\), is .......\(J\).

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Answer & Solution

Correct Answer

(B) \(32\)

Step-by-step Solution

Detailed explanation

\(W=\int \limits_0^4(2+3 x ) dx\)

\(=\left[2 x +\frac{3 x ^2}{2}\right]_0^4\)

\(=8+3 \times 8\)

\(=32\,J\)

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