Question

The potential energy of a particle of mass 1 kg moving in a plane is given by V= 3x + 4y, the position coordinates of the point being x and y. If the particle is at rest at (6,4) ; then find the magnitude of acceleration, speed at which it crosses the y-axis and the coordinate at which it cross the y-axis.

Answer 1

A particle of mass m = 1 kg moves in x-y plane, where there is a vector field E and vector force F.

      U = P.E. = 3 x + 4 y  Joules  in SI units
x component of force on the particle at (x,y):
      Fx = - dU/dx = - 3 Newtons
      Fy = - dU/dy = - 4 N
Magnitude of the force = 5 N

The direction of force on the particle at (x,y) = π + Tan⁻¹ (4/3)  with x axis or,   - (π - Tan⁻¹ 4/3).   The acceleration of the particle is in the same direction for all (x, y).

The displacement of the particle is along the path (straight line) with slope equal to :    tangent of that angle, ie., 4/3.

So the equation of path of the particle :  y = 4/3 x + c
As (6, 4) lies in this path:      4 = 4/3 * 6 +  c
                                              =>   c = - 4
=> equation of the path:  y = 4x/3 - 4    or   3 y - 4x + 12 = 0

This line meets y axis at x = 0, ie. ,  y = -4.
The point of crossing with y-axis:   (0, -4).

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Acceleration of the particle in the field:  5 m/sec^2 in the direction in the 3rd quadrant.

We can also find the velocity and energy of the particle as it crosses the y-axis.

The displacement between :  (6 ,4) and (0, -4) = 10 m
so  v² = u² + 2 a s
          = 0 + 2 * 5 * 10 = 100
       v = 10 m/s

Its KE = 1/2 * 1 kg * 10² = 50 J
Its PE = U = 3 x + 4 y = 3 * 0 - 4 * 4 = -16 J
         Total energy = 34 J