Question

Gravitational potential in a region varies as v(x,y)=(x^(2) +y^(2)). Find the force (ln N) experienced by 2kg mass placed at (3,4).​

Answer 1

Given:

Gravitational potential in a region varies as v(x,y)=(x^(2) +y^(2)).

To find:

Force experienced by a 2kg object placed at (3,4) ?

Calculation:

Gravitational potential is given as :

$\therefore \: V = {x}^{2} + {y}^{2}$

Now, Gravitational Potential Energy will be as follows:

$\therefore \: U= m \bigg \{ {x}^{2} + {y}^{2} \bigg \}$

$\implies \: U= 2\bigg \{ {x}^{2} + {y}^{2} \bigg \}$

$\implies \: U= 2 {x}^{2} +2 {y}^{2}$

Now, force along X axis:

$\therefore \: F_{x} = - \dfrac{ \partial U}{ \partial x}$

$\implies \: F_{x} = - \dfrac{ \partial \bigg(2 {x}^{2} + 2 {y}^{2} \bigg)}{ \partial x}$

$\implies \: F_{x} = - 4x$

$\implies \: F_{x} = - 4 \times 3$

$\implies \: F_{x} = - 12 \: N$

Again, force in Y axis:

$\therefore \: F_{y} = - \dfrac{ \partial U}{ \partial y}$

$\implies \: F_{y} = - \dfrac{ \partial \bigg(2 {x}^{2} + 2 {y}^{2} \bigg)}{ \partial y}$

$\implies \: F_{y} = - 4y$

$\implies \: F_{y} = - 4 \times 4$

$\implies \: F_{y} = - 16 \: N$

So, required force is:

$\therefore \: F_{net} = \sqrt{ {(F_{x})}^{2} + {(F_{y})}^{2} }$

$\implies \: F_{net} = \sqrt{{(-12)}^{2} + {(-16)}^{2} }$

$\implies \: F_{net} = \sqrt{144 + 256 }$

$\implies \: F_{net} = \sqrt{400 }$

$\implies \: F_{net} = 20 \: N$

So, net force is 20 N.