Question

A particle is moving on frictionless XY-plane. it
is acted on by a conservative force described by the
potential-energy function.

U(x, y, z) == ½ k(x² + y² + z²)

where, k= constant

Derive an expression for the force acting an the particle.​

Answer 1

Answer

Here we are provided with potential energy function U(x, y, z) = ½ k(x² + y² + z²) . U is function of x, y and z direction.

If work done by a force around a closed path is zero and it is independent of the path, such force is called conservative force.

Work done by the conservative force is stored in the form of potential energy.

Under conservative force : $\boxed{\sf F = - \dfrac{dU}{dx}}$

Now we have to find force acting on the particle in all three directions.

$\sf U_x = \dfrac{1}{2} K {x}^{2}$

$\sf U_y = \dfrac{1}{2} K {y}^{2}$

$\sf U_z = \dfrac{1}{2} K {z}^{2}$

Force acting on the particle in x - direction :

$\dashrightarrow \sf F(x) = - \dfrac{dU}{dx}$

$\dashrightarrow \sf F(x) = - \dfrac{d}{dx} \bigg [ \dfrac{1}{2} Kx^2 \bigg]$

$\dashrightarrow \sf F(x) = - Kx$

similarly,

Force acting on the particle in y - direction :

$\dashrightarrow \sf F(y) = - \dfrac{dU}{dy}$

$\dashrightarrow \sf F(y) = - \dfrac{d}{dy} \bigg [ \dfrac{1}{2} Ky^2 \bigg]$

$\dashrightarrow \sf F(y) = - Ky$

Force acting on the particle in z - direction :

$\dashrightarrow \sf F(z) = - \dfrac{dU}{dz}$

$\dashrightarrow \sf F(z) = - \dfrac{d}{dz} \bigg [ \dfrac{1}{2} Kz^2 \bigg]$

$\dashrightarrow \sf F(z) = - Kz$

Thus, the force acting an the particle :

$\sf \vec{F} = -K(x \hat{i} + y \hat{j} + z \hat{k})$

here k is constant so,

$\boxed{\sf \vec{F} = (x \hat{i} + y \hat{j} + z \hat{k})}$

Answer 2

I hope it helps all.

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