JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(...\,{J}.\)
- A \(300\)
- B \(75\)
- C \(150\)
- D \(450\)
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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