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**Hint:**The relation between force and potential energy is $F=-\dfrac{dU}{dr}$.When we find force along x-direction then we differentiate the potential energy with respect to $x$ and when we find the force along $y$-direction then we differentiate the potential energy with respect to $y$.

**Complete step by step answer:**

Spring force is a type of conservative force

$F=-\dfrac{dU}{dr}$

where, $U$is potential energy and $r$ is position vector in the direction of conservative force $F$ is conservative force.

Work done by the conservative force is equal to negative of change in potential energy

$w=-\Delta U$

Given, $U=\dfrac{1}{2}k{{x}^{2}}$

Put in the formula of force. When we calculate the force associated with this potential energy in x-direction then we take the position vector as $x$. It means that we differentiate the potential energy with respect to $x$.

$F=-\dfrac{dU}{dx}$

After putting the value of $U$in above equation then we get

$F=-\dfrac{d}{dx}(\dfrac{1}{2}k{{x}^{2}})$

$\Rightarrow F=-\dfrac{1}{2}k\dfrac{d}{dx}({{x}^{2}})$

The simple differentiation formula in the above equation is $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$.

$F=-\dfrac{1}{2}k\times 2x$

$\Rightarrow F=-kx$

Hence, the force along x-direction.

When we calculate the force associated with this potential energy in y-direction then we take the position vector as y it means that we differentiate the potential energy with respect to y.

$F=-\dfrac{dU}{dy}$

After putting the value of $U$in above equation then we get

$F=-\dfrac{d}{dy}(\dfrac{1}{2}k{{x}^{2}})$

$\Rightarrow F=-\dfrac{1}{2}k\dfrac{d}{dy}({{x}^{2}})$

When we differentiate ${{x}^{2}}$with respect to y then it became zero

$\therefore F=0$

**Hence, the force along y-direction is zero.**

**Note:**The negative sign shows the spring force is a type of restoring force. The elastic potential energy of an undeformed spring is taken to be zero. Spring force is a conservative force and the conservative force does not dissipate the energy of the system into heat.

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