The potential energy of a particle of mass 1 kg in a conservative field is given as `U = (3x^(2)y^(2) + 6x) J`, where x and y are measured in meter. Initially particle is at (1, 1) & at rest, then incorrect statement is :

Options

A. `vec(F) = - [(6xy^(2) + 6) hat(i) + 6x^(2)y hat(j)]`
B. Magnitude of initial acceleration of particle is `6 sqrt5 s^(2)`.
C. Work done to slowly bring the particle to origin is `- 9 J`
D. If particle is left free it move in stringht line.

Answer

D

Explanation

Force on the particle will be given as `vec(F) = bar(V) U`
`= (delU)/(delx) hat(i) - (del U)/(del y) = - (6 xy^(2) + 6) hat(i) - (6x^(2) y) hat(j)`
Now, for acceleration at `( t = 0)` and `x = 1, y = 1`
`vec(a) = (vec(F))/(M) = 12 hat(i) - 6 hat(j)`
`|vec(a)| = .^(6)sqrt5 m//s^(2)`
Particle is at rest at `x = 1, y = 1`, then
`P.E. + K.E. = M.E.`
`u (1, 1) + K.E. (1, 1) = M.E.`
`rArr M.E. = 9 J`
`DeltaU + DeltaK = DeltaW`
`DeltaW = DeltaU = U (0, 0) - U (1, 1)`
`DeltaW = -9 J`