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Question: The potential energy function of a particle in a region of space is given as \( U = (2{x^2} + 3{y^3}...

The potential energy function of a particle in a region of space is given as U=(2x2+3y3+2x)JU = (2{x^2} + 3{y^3} + 2x)J . Here x, y, z is in meters. Find the force acting on the article at point P (1m,2m,3m).

Explanation

Solution

Hint
We will find the force by partial differentiation of potential energy with respect to the position. Then we will substitute the values of x, y and z respectively to get the desired answer.

Complete step by step answer
F is the force exerted by the force field such as gravity, spring force etc. U is the potential energy equal to work that must be done against the force to move an object from reference point (U=0)\left( {U = 0} \right) to the distance r.
This definition can also be written as U=F.drU = - \int {F.dr}
F=dUdrF = - \dfrac{{dU}}{{dr}}
It can also be written as F=δUδxiδUδyjδUδzkF = - \dfrac{{\delta U}}{{\delta x}}i - \dfrac{{\delta U}}{{\delta y}}j - \dfrac{{\delta U}}{{\delta z}}k where i ,j and k are the unit vectors along the axes.
F=δ(2x2+2x)δxiδ(3y3)δyjF = - \dfrac{{\delta (2{x^2} + 2x)}}{{\delta x}}i - \dfrac{{\delta (3{y^3})}}{{\delta y}}j , δanδa=nan1\dfrac{{\delta {a^n}}}{{\delta a}} = n{a^{n - 1}}
F=(4x+2)i(9y2)jF = - (4x + 2)i - (9{y^2})j
Here (X=1m)(X = 1m) and y=2my = 2m
F=6i36jF = - 6i - 36j N
F=6(i+6j)F = - 6(i + 6j) N.

Additional Information
We can define potential energy for any conservative force. Conservative forces are those forces in work done by it or against it only depends on the starting and finishing point. It doesn’t depend on the path followed.

Note
You have to do the partial differentiation instead of simple differentiation of potential energy. Here the Force F is in vector form so do not simply add the components.