
The potential energy of spring is given by $U=\dfrac{1}{2}k{{x}^{2}}$, where $x$ is extension spring. Find the force associated with this potential energy in y-direction.
Answer
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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.
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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