Question

A particle in a certain conservative force field has a potential energy given by$U = \dfrac{{20xy}}{z}$. The force exerted on it is:
A. $\left( {\dfrac{{20y}}{z}} \right)\hat i + \left( {\dfrac{{20x}}{z}} \right)\hat j + \left( {\dfrac{{20xy}}{{{z^2}}}} \right)\hat k$
B. $- \left( {\dfrac{{20y}}{z}} \right)\hat i - \left( {\dfrac{{20x}}{z}} \right)\hat j + \left( {\dfrac{{20xy}}{{{z^2}}}} \right)\hat k$
C. $- \left( {\dfrac{{20y}}{z}} \right)\hat i - \left( {\dfrac{{20x}}{z}} \right)\hat j - \left( {\dfrac{{20xy}}{{{z^2}}}} \right)\hat k$
D. $\left( {\dfrac{{20y}}{z}} \right)\hat i + \left( {\dfrac{{20x}}{z}} \right)\hat j - \left( {\dfrac{{20xy}}{{{z^2}}}} \right)\hat k$

Answer 1

Hint:- The potential energy is the energy which an object attains at a particular position in its motion. The force due to potential energy is the force required to move the object from the reference point to a position which is at a distance r from the reference point.
Formula used: The formula of the force exerted by a particle in conservative field having a potential energy is given by,
$F = - \nabla U$
Where $\nabla$ is equal to $\nabla = \dfrac{\partial }{{\partial x}}\hat i + \dfrac{\partial }{{\partial y}}\hat j + \dfrac{\partial }{{\partial z}}\hat k$ and $U$ is the potential energy of the particle. Also$\hat i$,$\hat j$ and $\hat k$ are directions representing x-direction ,y-direction and z-direction.

Complete step-by-step solution
It is given that the potential energy of a particle is equal to $U = \dfrac{{20xy}}{z}$ and we have to find the force that is exerted on the particle.
As the force exerted on the particle is given by,
$F = - \nabla U$
Where $\nabla$ is equal to $\nabla = \dfrac{\partial }{{\partial x}}\hat i + \dfrac{\partial }{{\partial y}}\hat j + \dfrac{\partial }{{\partial z}}\hat k$ and $U$ is the potential energy of the particle.
Therefore, the force is given by,
$\Rightarrow F = - \nabla U$
Replace the value of potential energy in the above equation and the differentiating it partially.
$\Rightarrow F = - \nabla \left( {\dfrac{{20xy}}{z}} \right)$
$\Rightarrow F = - \left( {\dfrac{\partial }{{\partial x}}\hat i + \dfrac{\partial }{{\partial y}}\hat j + \dfrac{\partial }{{\partial z}}\hat k} \right) \cdot \left( {\dfrac{{20xy}}{z}} \right)$
After differentiating the potential energy we get,
$\Rightarrow F = - \left( {\dfrac{{20y}}{z}\hat i + \dfrac{{20x}}{z}\hat j - \dfrac{{20xy}}{{{z^2}}}\hat k} \right)$
Solving furthermore we get,
$\Rightarrow F = - \dfrac{{20y}}{z}\hat i - \dfrac{{20x}}{z}\hat j + \dfrac{{20xy}}{{{z^2}}}\hat k$.
The force applied on the particle is given by$F = - \dfrac{{20y}}{z}\hat i - \dfrac{{20x}}{z}\hat j + \dfrac{{20xy}}{{{z^2}}}\hat k$. The correct answer for this problem is option B.

Note:- It is important for students to differentiate the potential energy with respect to x, y and z with care as it is not a normal process of differentiation but this is the partial differentiation of the potential energy. The partial differential is done such that if a given term is differentiated with respect to x then every term except x is taken as constant.