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The potential energy for force field F is given by U(x,y) = cos(x+y). The force acting on a particle at position given by coordinates ( 0 , pie/4) is

muktesh singh , 11 Years ago
Grade 11
anser 1 Answers
Askiitians Tutor Team

To determine the force acting on a particle at the coordinates (0, π/4) given the potential energy function \( U(x, y) = \cos(x + y) \), we first need to understand the relationship between potential energy and force. The force can be derived from the potential energy using the negative gradient of the potential energy function.

Understanding the Gradient

The force \( \mathbf{F} \) in a conservative force field is related to the potential energy \( U \) by the equation:

F = -∇U

Here, \( ∇U \) (the gradient of U) is a vector that points in the direction of the greatest rate of increase of the potential energy, and its magnitude is the rate of increase in that direction. The negative sign indicates that the force acts in the direction of decreasing potential energy.

Calculating the Gradient

To find the gradient \( ∇U \), we need to compute the partial derivatives of \( U \) with respect to \( x \) and \( y \):

  • Partial derivative with respect to x:
  • We differentiate \( U(x, y) = \cos(x + y) \) with respect to \( x \):

    ∂U/∂x = -sin(x + y)

  • Partial derivative with respect to y:
  • Next, we differentiate \( U(x, y) \) with respect to \( y \):

    ∂U/∂y = -sin(x + y)

Thus, the gradient is:

∇U = (-sin(x + y), -sin(x + y))

Evaluating the Gradient at (0, π/4)

Now, we substitute the coordinates (0, π/4) into the gradient:

First, calculate \( x + y \):

0 + π/4 = π/4

Now, we find the sine of \( π/4 \):

sin(π/4) = √2/2

Substituting this into the gradient gives:

∇U(0, π/4) = (-√2/2, -√2/2)

Finding the Force

Now, we can find the force by applying the negative gradient:

F(0, π/4) = -∇U(0, π/4) = (√2/2, √2/2)

Final Result

Therefore, the force acting on the particle at the position (0, π/4) is:

F = (√2/2, √2/2)

This means that the force has equal components in both the x and y directions, pointing towards the direction of decreasing potential energy in the field defined by \( U(x, y) \).

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