The potential energy function associated with the force $¯ F = 4 x y^i + 2 x^{2}^j$ is:

U = - 4

x^{2} y

+ constant

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

U =

x^{2} y

+ constant

No worries! We‘ve got your back. Try BYJU‘S free classes today!

U =

x^{3} y

+ constant

No worries! We‘ve got your back. Try BYJU‘S free classes today!

not defined

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A U = - 4 $x^{2} y$ + constant

Given,

Force, $F = 4 x y^i + 2 x^{2}^j$

Position vector, $\to R = x^i + y^j$

Potential energy = -work done= $- \int \to F . d \to R$

$= - \int (4 x y^i + 2 x^{2}^j) . (d x^i + d y^j)$

$= - \int 4 x y d x - \int 2 x^{2} d y$

$= - 2 x^{2} y - 2 x^{2} y + c$

$= - 4 x^{2} y + c$

Hence, potential energy Is $- 4 x^{2} y + c$

Suggest Corrections

Similar questions

{(\begin{matrix} \frac{x^{- 1} y^{2}}{x^{2} y^{- 4}} \end{matrix})}^{7} \div {(\begin{matrix} \frac{x^{3} y^{- 5}}{x^{- 2} y^{3}} \end{matrix})}^{- 6} = x^{p} . y^{q}

p + q =

x^{2} y + x y^{2}

x^{3} y + x z + x y^{2}

(\frac{x^{- 1} y^{2}}{x^{2} y^{- 4}})^{7} \div (\frac{x^{3} y^{- 5}}{x^{- 2} y^{3}})^{- 5} = x^{p} . y^{q}

p + q

x y^{2}

= x^{2} y

¯ F

U (x, y) = s i n (x + y)

m

(0, \frac{π}{4})

Join BYJU'S Learning Program