A particle of mass `m` moves with a variable velocity v, which changes with distance covered x along a straight line as `v=ksqrtx`, where k is a positive constant. The work done by all the forces acting on the particle, during the first t seconds is

Options

A. `(mk^(4)t^(2))/(4)`
B. `(mk^(2)t)/(2)`
C. `(ml^(4)t^(2))/(8)`
D. `(mk^(2)t^(2))/(4)`

Answer

To find the work done by all the forces acting on a particle of mass \( m \) moving with a variable velocity \( v = k \sqrt{x} \), we can follow these steps:

### Step 1: Relate velocity to distance
Given the velocity \( v \) is defined as:
\[ v = k \sqrt{x} \]
We can express this in terms of the differential relationship:
\[ v = \frac{dx}{dt} \]
Thus, we have:
\[ \frac{dx}{dt} = k \sqrt{x} \]

### Step 2: Rearranging the equation
Rearranging gives:
\[ \frac{dx}{\sqrt{x}} = k \, dt \]

### Step 3: Integrating both sides
Now, we integrate both sides. The left side requires a substitution:
\[ \int \frac{dx}{\sqrt{x}} = \int k \, dt \]
The integral of the left side is:
\[ 2\sqrt{x} = kt + C \]
where \( C \) is the constant of integration.

### Step 4: Determine the constant of integration
At \( t = 0 \), \( x = 0 \):
\[ 2\sqrt{0} = k(0) + C \implies C = 0 \]
Thus, we have:
\[ 2\sqrt{x} = kt \]

### Step 5: Solve for \( x \)
Squaring both sides gives:
\[ 4x = k^2 t^2 \implies x = \frac{k^2 t^2}{4} \]

### Step 6: Substitute back to find velocity
Now, substituting \( x \) back into the equation for \( v \):
\[ v = k \sqrt{x} = k \sqrt{\frac{k^2 t^2}{4}} = \frac{k^2 t}{2} \]

### Step 7: Calculate the change in kinetic energy
The work done \( W \) is equal to the change in kinetic energy. The initial kinetic energy \( KE_i \) is zero (since the initial velocity is zero):
\[ KE_i = 0 \]
The final kinetic energy \( KE_f \) is given by:
\[ KE_f = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{k^2 t}{2}\right)^2 \]
Calculating this:
\[ KE_f = \frac{1}{2} m \cdot \frac{k^4 t^2}{4} = \frac{m k^4 t^2}{8} \]

### Step 8: Work done
Thus, the work done \( W \) is:
\[ W = KE_f - KE_i = \frac{m k^4 t^2}{8} - 0 = \frac{m k^4 t^2}{8} \]

### Final Answer
The work done by all the forces acting on the particle during the first \( t \) seconds is:
\[ W = \frac{m k^4 t^2}{8} \]

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