The potential energy `varphi`, in joule, of a particle of mass `1kg`, moving in the x-y plane, obeys the law `varphi=3x+4y`, where `(x,y)` are the coordinates of the particle in metre. If the particle is at rest at `(6,4)` at time `t=0`, then

Options

A. The particle has constant acceleration
B. The work done by external forces from the position of rest to the instant the particle crossing x-axis is `25J`
C. The speed of the particle when it crosses the y-axis is `10ms^(-1)`
D. The coordinates of the particle at time `t = 4s` are `(-18, -28)` (in m)

Answer

A, B, C, D

Explanation

`phi = 3 x + 4y`
`vec(F) = (- delta phi)/(delta x) hat(i) - (delta phi)/(delta y) hat(j)`
`vec(F) = - 3 hat(i) - 4 hat(j)`
Potential energy at (6, 4)
`phi = 3 (6) + 4 (4) = 18 + 16 = 34` Joule
Here force is conservative so its work done will not depends upon the path choosen.
(A) acceleration of particle
`a = (F)/(m) rArr a = (-3 hat(i) - 4 hat(j)) m//sec^(2)`. (It is constant)
`vec(u) = 0 hat(i) + 0 hat(j)`
`t = 4` sec
`s_(x) = (s_(x))_(0) + u_(x) t + (1)/(2) a_(x) t^(2)`
`s_(x) = 6 + 0 - (1)/(2) (3) xx (16) rArr s_(x) = 6 - 24 rArr (s_(x) = - 18)`
`s_(y) = (s_(y))_(0) + u_(y) t (1)/(2) a_(y) t^(2)`
`rArr s^(y) = 4 + 0 - (1)/(2) (4) (16)`
`rArr s_(y) = 4 + 0 - (1)/(2) (4) (16)`
`rArr s^(y) = 4 - 32`
`rArr (s_(y) = - 28)`
(C) Speed of particle when it crosses y - axis i.e. at that time its x - coordinate will be zero.
`s_(x) = (s_(x))_(0) + a_(x) t + (1)/(2) a_(x) t^(2)`
` 0 = 6 + 0 - (1)/(2) (3) (t)^(2) rArr (3)/(2) t^(2) = 6`
`rArr t = 2` sec
`v_(x) = u_(x) + a_(x) t rArr v_(x) = - 3 (2) rArr v_(x) = - 6 m//sec`
`v_(y) = u_(y) + a_(y) t rArr v_(y) = 0 - 4 (2) rArr v_(y) = - 8 m//sec`
Speed of particle `= sqrt((-6)^(2) + (-8)^(2)) = 10 m//sec`
(B) When the body is crossing x-axis than that time `s_(y) = 0`
`s_(y) = (s_(y))_(0) + t + (1)/(2) xx a_(y) t^(2)`
`0 = 4 + 0 + (1)/(2) xx (-4) t^(2)`
`2t^(2) = 4 rArr t = sqrt2 sec`
`v_(x) = a_(x) t rArr v_(x) = - 3 sqrt2 m//sec`
`v_(y) = - 4 sqrt2 m//sec`
`KE = (1)/(2) mv_(x)^(2) + (1)/(2) mv_(y)^(2)`
`KE = (1)/(2) xx 1 xx (3 sqrt2)^(2) + (4 sqrt2)^(2)`
`rArr KE = 25` Joules