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Work Power and Energy question

2024 · 29 Jan · Shift 1 · Q79
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  5. /2024 · 29 Jan · Shift 1 · Q79

Work Power and Energy question

2024 · 29 Jan · Shift 1 · Q79

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
The potential energy function (in JJJ) of a particle in a region of space is given as U=(2x2+3y3+2z)U=\left(2 x^2+3 y^3+2 z\right)U=(2x2+3y3+2z). Here x,yx, yx,y and zzz are in meter. The magnitude of xxx-component of force (in NNN) acting on the particle at point P(1,2,3)mP(1,2,3) \mathrm{m}P(1,2,3)m is :
  1. A
    4
  2. B
    2
  3. C
    8
  4. D
    6
View written solutionFree

Correct answer: A

  1. The force is related to potential energy by F⃗=−∇U\vec F=-\nabla UF=−∇U So the xxx-component of force is Fx=−∂U∂xF_x=-\frac{\partial U}{\partial x}Fx​=−∂x∂U​

  2. Given U=2x2+3y3+2zU=2x^2+3y^3+2zU=2x2+3y3+2z Differentiate with respect to xxx: ∂U∂x=4x\frac{\partial U}{\partial x}=4x∂x∂U​=4x Hence, Fx=−4xF_x=-4xFx​=−4x

  3. At point P(1,2,3)P(1,2,3)P(1,2,3), we have x=1x=1x=1. Therefore, Fx=−4(1)=−4 NF_x=-4(1)=-4\,\text{N}Fx​=−4(1)=−4N

  4. The question asks for the magnitude of the xxx-component: ∣Fx∣=4 N|F_x|=4\,\text{N}∣Fx​∣=4N

  5. Therefore, the correct option is: A: 4\boxed{\text{A: }4}A: 4​

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