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 .......\(J\).
Pick an option to check your answer.
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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