The potential energy (in joules) of body of mass 2 kg moving inXY plane is given by U= 6x+8y, where x and y are in metres.If body is at rest point(6m,4m) at time t=0,it will cross y- axis at time t equal to?

To determine when the body crosses the y-axis, we first need to analyze the given potential energy function and the conditions of the problem. The potential energy \( U \) is defined as \( U = 6x + 8y \), where \( x \) and \( y \) are the coordinates of the body in the XY plane. Since the body is at rest at the point (6 m, 4 m) at time \( t = 0 \), we can infer that it will start moving due to the forces acting on it as it transitions to a different position.

Understanding the Motion

The potential energy function indicates that the forces acting on the body can be derived from the potential energy. The force \( \mathbf{F} \) can be calculated using the negative gradient of the potential energy:

Force Calculation:

• In the x-direction: \( F_x = -\frac{\partial U}{\partial x} = -6 \)
• In the y-direction: \( F_y = -\frac{\partial U}{\partial y} = -8 \)

This means the body experiences a constant force of \( -6 \, \text{N} \) in the x-direction and \( -8 \, \text{N} \) in the y-direction. Since these forces are constant, the body will accelerate in both directions.

Calculating Acceleration

Using Newton's second law, we can find the acceleration:

• Acceleration in x-direction: \( a_x = \frac{F_x}{m} = \frac{-6}{2} = -3 \, \text{m/s}^2 \)
• Acceleration in y-direction: \( a_y = \frac{F_y}{m} = \frac{-8}{2} = -4 \, \text{m/s}^2 \)

Equations of Motion

Since the body starts from rest, we can use the equations of motion to find its position as a function of time:

Position Equations:

• For x: \( x(t) = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 \)
• For y: \( y(t) = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 \)

Given that the initial position \( (x_0, y_0) = (6, 4) \) and the initial velocities \( v_{0x} = 0 \) and \( v_{0y} = 0 \), we can simplify these equations:

• For x: \( x(t) = 6 - \frac{1}{2} \cdot 3 t^2 = 6 - 1.5t^2 \)
• For y: \( y(t) = 4 - \frac{1}{2} \cdot 4 t^2 = 4 - 2t^2 \)

Finding the Time to Cross the Y-Axis

The body crosses the y-axis when \( x(t) = 0 \). Setting the equation for \( x(t) \) to zero gives:

0 = 6 - 1.5t^2

1.5t^2 = 6

t^2 = \frac{6}{1.5} = 4

t = 2 \, \text{s}

Final Position on the Y-Axis

Now, we can find the corresponding y-coordinate when \( t = 2 \):

y(2) = 4 - 2(2^2) = 4 - 8 = -4

Thus, the body crosses the y-axis at the point (0, -4) at time \( t = 2 \, \text{s} \).

In summary, the body will cross the y-axis at time \( t = 2 \, \text{s} \), moving from its initial position under the influence of the forces derived from its potential energy. This analysis illustrates how potential energy can be linked to motion through forces and accelerations in a systematic way.