JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
The potential energy function (in J) of a particle in a region of space is given as \(U=\left(2 x^2+3 y^3+2 z\right)\). Here \(\mathrm{x}, \mathrm{y}\) and \(\mathrm{z}\) are in meter. The magnitude of \(\mathrm{x}\) - component of force (in \(\mathrm{N}\) ) acting on the particle at point \(\mathrm{P}(1,2,3)\ \mathrm{m}\) is:
- A \(2\)
- B \(6\)
- C \(4\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
Given \(\mathrm{U}=2 \mathrm{x}^2+3 \mathrm{y}^3+2 \mathrm{z}\) \(\mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=-4 \mathrm{x}\) At \(x=1\) magnitude of \(F_x\) is \(4 \mathrm{~N}\)
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