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
- A 201 J
- B 215 J
- C 229 J
- D 307 J
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, \(W=\int_{\mathrm{x}_1}^{\mathrm{x}_2} F d x=\int_2^5\left(6 \mathrm{x}^2-4 \mathrm{x}+3\right) \mathrm{dx}\) \(=\left[6 \cdot \frac{x^3}{3}-4 \frac{x^2}{2}+3 x\right]_2^5=201 J\)
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