A particle of mass m initially at rest. A variabl force acts on the particle `f=kx^(2)` where k is a constant and x is the displacment. Find the work done by the force f when the speed of particles is v.

Options

A. `1/2kx^(2)`
B. `1/2kv^(2)`
C. `1/2mv^(2)`
D. `2mv^(2)`

Answer

To solve the problem step by step, we will follow the outlined reasoning in the video transcript and derive the work done by the variable force \( f = kx^2 \) when the speed of the particle is \( v \).

### Step 1: Understand the Given Information
- The mass of the particle is \( m \).
- The force acting on the particle is given by \( f = kx^2 \), where \( k \) is a constant and \( x \) is the displacement.
- The particle starts from rest, so the initial velocity \( u = 0 \).
- We need to find the work done by the force when the speed of the particle is \( v \).

**Hint:** Identify the initial conditions and the relationship between force, mass, and acceleration.

### Step 2: Relate Force to Acceleration
Using Newton's second law, we can write:
\[ f = ma \]
Substituting the expression for force:
\[ kx^2 = ma \]
From this, we can express acceleration \( a \) as:
\[ a = \frac{kx^2}{m} \]

**Hint:** Recall that acceleration can also be expressed as the derivative of velocity with respect to time.

### Step 3: Express Acceleration in Terms of Velocity
We know that acceleration \( a \) can also be expressed as:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]
Setting the two expressions for acceleration equal gives:
\[ \frac{kx^2}{m} = v \frac{dv}{dx} \]

**Hint:** This step involves using the chain rule of calculus to relate acceleration, velocity, and displacement.

### Step 4: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{kx^2}{m} dx = v dv \]

**Hint:** This is a differential equation that can be integrated to find the relationship between \( x \) and \( v \).

### Step 5: Integrate Both Sides
Integrating both sides, we have:
\[ \int \frac{kx^2}{m} dx = \int v dv \]
The left side integrates to:
\[ \frac{kx^3}{3m} \]
The right side integrates to:
\[ \frac{v^2}{2} \]
Thus, we have:
\[ \frac{kx^3}{3m} = \frac{v^2}{2} \]

**Hint:** Make sure to set the limits of integration correctly based on the initial and final conditions.

### Step 6: Solve for \( x^3 \)
Rearranging gives:
\[ kx^3 = \frac{3mv^2}{2} \]
So,
\[ x^3 = \frac{3mv^2}{2k} \]

**Hint:** This expression will help us find the work done by the force.

### Step 7: Calculate the Work Done
The work done \( W \) by the force is given by:
\[ W = \int F \, dx = \int kx^2 \, dx \]
Evaluating this integral from \( 0 \) to \( x \):
\[ W = \left[ \frac{kx^3}{3} \right]_{0}^{x} = \frac{kx^3}{3} \]
Substituting \( x^3 \) from the previous step:
\[ W = \frac{k}{3} \left( \frac{3mv^2}{2k} \right) = \frac{mv^2}{2} \]

**Hint:** The work done can be expressed in terms of the final speed of the particle.

### Final Result
Thus, the work done by the force when the speed of the particle is \( v \) is:
\[ W = \frac{mv^2}{2} \]

**Conclusion:** The work done by the force \( f \) when the speed of the particle is \( v \) is \( \frac{mv^2}{2} \).