The potential energy of a particle of mass 5 kg moving in the `x-y` plane is given by `U=(-7x+24y)J`, where x and y are given in metre. If the particle starts from rest, from the origin, then the speed of the particle at `t=2`s is

Options

A. 5 m/s
B. 14 m/s
C. 17.5 m/s
D. 10 m/s

Answer

To solve the problem, we need to determine the speed of a particle of mass 5 kg moving in the x-y plane, given its potential energy function \( U = -7x + 24y \) J. The particle starts from rest at the origin and we need to find its speed at \( t = 2 \) seconds.

### Step-by-Step Solution:

1. **Identify the Potential Energy Function**:
The potential energy \( U \) of the particle is given as:
\[ U = -7x + 24y \]

2. **Calculate the Force**:
The force acting on the particle can be found using the relation:
\[ \vec{F} = -\nabla U \]
where \( \nabla U \) is the gradient of the potential energy. We calculate the components of the force:
- In the x-direction:
\[ F_x = -\frac{dU}{dx} = -(-7) = 7 \, \text{N} \]
- In the y-direction:
\[ F_y = -\frac{dU}{dy} = -24 \, \text{N} \]
Thus, the force vector is:
\[ \vec{F} = (7 \hat{i} - 24 \hat{j}) \, \text{N} \]

3. **Calculate the Acceleration**:
Using Newton's second law, \( \vec{F} = m \vec{a} \), we can find the acceleration:
- For the x-component:
\[ a_x = \frac{F_x}{m} = \frac{7}{5} = 1.4 \, \text{m/s}^2 \]
- For the y-component:
\[ a_y = \frac{F_y}{m} = \frac{-24}{5} = -4.8 \, \text{m/s}^2 \]
Therefore, the acceleration vector is:
\[ \vec{a} = (1.4 \hat{i} - 4.8 \hat{j}) \, \text{m/s}^2 \]

4. **Determine the Velocity Components**:
The particle starts from rest, so the initial velocities \( v_{0x} = 0 \) and \( v_{0y} = 0 \). We can use the kinematic equation:
\[ v = u + at \]
- For the x-component:
\[ v_x = 0 + (1.4)(2) = 2.8 \, \text{m/s} \]
- For the y-component:
\[ v_y = 0 + (-4.8)(2) = -9.6 \, \text{m/s} \]

5. **Calculate the Magnitude of the Velocity**:
The speed of the particle is given by the magnitude of the velocity vector:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(2.8)^2 + (-9.6)^2} \]
Calculating this:
\[ v = \sqrt{7.84 + 92.16} = \sqrt{100} = 10 \, \text{m/s} \]

### Final Answer:
The speed of the particle at \( t = 2 \) seconds is \( 10 \, \text{m/s} \).

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