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NEET · Physics · STD 11 - 5. work,energy,power and collision

A force \(\mathrm{F}=20+10 \mathrm{y}\) acts on a particle in \(y\) direction where \(\mathrm{F}\) is in newton and \(\mathrm{y}\) in meter. Work done by this force to move the particle from \(y=0\) to \(y=1 \;\mathrm{m}\) is......\(J\)

  1. A \(30\)
  2. B \(5\)
  3. C \(25\)
  4. D \(20\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(25\)

Step-by-step Solution

Detailed explanation

\(\mathrm{W}=\int_{\mathrm{y}_{\mathrm{1}}}^{y_{\mathrm{2}}} \mathrm{F} \mathrm{dy}\) \(\Rightarrow \mathrm{W}=\int_{0}^{1}(20+10 \mathrm{y}) \mathrm{dy}\)
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