The potential energy of a particle of mass 5 kg moving in the x-y plane is given by `U=-7x+24y` joule, x and y being in metre. Initially at t = 0 the particle is at the origin. (0, 0) moving with a velocity of `6[2.4hat(i)+0.7hat(j)]m//s`. The magnitude of force on the particle is :

Options

A. 25 unit
B. 24 unit
C. 7 unit
D. None of these

Answer

To find the magnitude of the force acting on the particle, we can follow these steps:

### Step 1: Identify the given potential energy function
The potential energy \( U \) of the particle is given by:
\[ U = -7x + 24y \quad \text{(in joules)} \]

### Step 2: Calculate the partial derivatives of the potential energy
To find the force, we need to calculate the partial derivatives of the potential energy with respect to \( x \) and \( y \).

1. **Partial derivative with respect to \( x \)**:
\[ \frac{\partial U}{\partial x} = -7 \]
(The derivative of \(-7x\) is \(-7\), and the derivative of \(24y\) with respect to \(x\) is \(0\) since \(y\) is treated as a constant.)

2. **Partial derivative with respect to \( y \)**:
\[ \frac{\partial U}{\partial y} = 24 \]
(The derivative of \(-7x\) with respect to \(y\) is \(0\), and the derivative of \(24y\) is \(24\).)

### Step 3: Use the force formula
The force \( \mathbf{F} \) acting on the particle in the x-y plane can be expressed as:
\[ \mathbf{F} = -\frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} \]
Substituting the values we found:
\[ \mathbf{F} = -(-7) \hat{i} + 24 \hat{j} = 7 \hat{i} + 24 \hat{j} \]

### Step 4: Calculate the magnitude of the force
The magnitude of the force \( |\mathbf{F}| \) is given by:
\[ |\mathbf{F}| = \sqrt{(F_x)^2 + (F_y)^2} \]
Substituting \( F_x = 7 \) and \( F_y = 24 \):
\[ |\mathbf{F}| = \sqrt{(7)^2 + (24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ N} \]

### Final Answer
The magnitude of the force on the particle is:
\[ \boxed{25 \text{ N}} \]