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

Answer

A, B, C, D

Explanation

Since potential energy of the particle is equal to `phi`, x-component of the force acting on the particle is equal to
`F_x=-(dphi)/(dx)`
and y-component of the force on the particle is equal to
`F_y=-(dphi)/(dy)`
Hence, force acting on the particle is `vecF=-(3hati+4hatj)N`
It means, force acting on the particle is constant. Hence, the particle moves with constant acceleration. So option(a) is correct.
Acceleration of the particle, `veca=(vecF)/(m_2)=-(3hati+4hatj)ms^-1`
Its magnitude is `a=sqrt(3^2+4^2)=5ms^-2`
Since, the particle was initially at rest at `(6,4)`, position of the particle at time t is given by
`x=6+1/2a_xt^2=(6-3/2t^2)m`
and `y=4+1/2a_yt^2=(4-2t^2)m`
when the particle crosses the x-axis, `y=0`
`t_1=sqrt2s`
Displacement of the particle during this time,
`s_1=1/2at^2=5m`
Hence, work done by the force, up to this instant
`Fs_1=5xx5J=25J`
Hence, obtion b. is also correct.
The particle crosses y-axis when `x=0`
Hence, `6-3/2t_2^2=0` or `t_2=2s`
Speed of the particle at this instant will be
`v=at_2=5xx2=10ms^-1`
Hence, option c. is also correct.
At `t=4s` , `x=6-3/2(4)^2=-18m`
and `y=4-2(4)^2=-28m`
Hence, option (d) is also correct.