JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2021
- Medium
- Numerical
A force of \(F=(5 y+20) \hat{j} \,N\) acts on a particle. The work done by this force when the particle is moved from \(y=0 \,m\) to \(y=10 \,{m}\) is \(...\,{J}.\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(450\)
Step-by-step Solution
Detailed explanation
\(F=(5 y+20) \hat{j}\)
\(W=\int F d y=\int_{0}^{10}(5 y+20) d y\)
\(=\left(\frac{5 y^{2}}{2}+20 y\right)_{0}^{10}\)
\(=\frac{5}{2} \times 100+20 \times 10\)
\(W=250+200=450 J\)
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