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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Application +4 | -1
The potential energy of a body is given by

U=40+6x2−7xy+8y2+32zU=40+6 x^2-7 x y+8 y^2+32 z
where UU is in joule and x,y,zx, y, z in metre. Deduce the x,yx, y and zz components of the force (in newton) on the body when it is in position (−2,0,+5)(-2,0,+5)
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Correct option: D
Fx=24;Fy=−14;Fz=−32F_x=24 ; F_y=-14 ; F_z=-32
Option D (Fx=24;Fy=−14;Fz=−32F_x=24 ; F_y=-14 ; F_z=-32) is correct because it's calculated by taking the negative partial derivatives of the potential energy with respect to x, y, and z.

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