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. `5ms^(-1)`
B. `14ms^(-1)`
C. `17.5ms^(-1)`
D. `10ms^(-1)`

Answer

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Identify Given Information
- Mass of the particle, \( m = 5 \, \text{kg} \)
- Potential energy, \( U = -7x + 24y \, \text{J} \)
- Initial velocity, \( u = 0 \, \text{m/s} \) (since the particle starts from rest)
- Time, \( t = 2 \, \text{s} \)

### Step 2: Determine the Force from Potential Energy
The force acting on the particle can be derived from the potential energy using the relation:
\[ \vec{F} = -\nabla U \]
This means we need to calculate the partial derivatives of \( U \) with respect to \( x \) and \( y \).

#### Calculate \( F_x \):
\[ F_x = -\frac{dU}{dx} = -\frac{d}{dx}(-7x + 24y) = 7 \, \text{N} \]

#### Calculate \( F_y \):
\[ F_y = -\frac{dU}{dy} = -\frac{d}{dy}(-7x + 24y) = -24 \, \text{N} \]

### Step 3: Calculate the Net Force
Now we can find the net force \( \vec{F} \):
\[ \vec{F} = (F_x, F_y) = (7, -24) \, \text{N} \]

### Step 4: Calculate the Magnitude of the Net Force
The magnitude of the net force can be calculated using the Pythagorean theorem:
\[ F_{\text{net}} = \sqrt{F_x^2 + F_y^2} = \sqrt{7^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \, \text{N} \]

### Step 5: Calculate the Acceleration
Using Newton's second law, we can find the acceleration \( a \):
\[ a = \frac{F_{\text{net}}}{m} = \frac{25 \, \text{N}}{5 \, \text{kg}} = 5 \, \text{m/s}^2 \]

### Step 6: Calculate the Final Velocity
Using the first equation of motion:
\[ v = u + at \]
Substituting the known values:
\[ v = 0 + (5 \, \text{m/s}^2)(2 \, \text{s}) = 10 \, \text{m/s} \]

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

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