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AP EAMCET · PHYSICS · Work Power Energy

A force of \(\left(6 x^2-4 x+3\right) \mathrm{N}\) acts on a body of mass 0.75 kg and displaces it from \(x=2 \mathrm{~m}\) to \(x=5 \mathrm{~m}\). The work done by the force is

  1. A 201 J
  2. B 215 J
  3. C 229 J
  4. D 307 J
Verified Solution

Answer & Solution

Correct Answer

(A) 201 J

Step-by-step Solution

Detailed explanation

\(F=\left(6 x^2-4 x+3\right) N, m=0.75 \mathrm{~kg}, x_1=2 m, x_2=5 \mathrm{~m}\) Work done, \(W=\int_{\mathrm{x}_1}^{\mathrm{x}_2} F d x=\int_2^5\left(6 \mathrm{x}^2-4 \mathrm{x}+3\right) \mathrm{dx}\) \(=\left[6 \cdot \frac{x^3}{3}-4 \frac{x^2}{2}+3 x\right]_2^5=201 J\)