Question

The potential energy U of a particle of mass m = 1 kg moving in x - y plane is given by U = 3x + 4y, where x and y are in metre and U is in joule. If initially particle was at rest, then its speed at t = 2 s will be
A 10 m/s
B 6 m/s
C 8 m/s
D 7 m/s​

Answer 1

Question:

The potential energy U of a particle of mass m = 1 kg moving in x - y plane is given by U = 3x + 4y, where x and y are in metre and U is in joule. If initially particle was at rest, then its speed at t = 2 s will be

A 10 m/s

B 6 m/s

C 8 m/s

D 7 m/s

Solution:

• 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):

1.   Fx = - dU/dx = - 3 Newton
2. Fy = - dU/dy = - 4 N
3. 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

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