A block is pulled with constant power what is the relation between displacement(x) and time(t) for this

Options

A. `x prop t^(3/2)`
B. `x prop t^(2/3)`
C. `x prop t^(1/2)`
D. `x prop t^2`

Answer

To find the relationship between displacement \( x \) and time \( t \) when a block is pulled with constant power, we can follow these steps:

### Step 1: Understand the relationship between power, force, and velocity
The power \( P \) can be expressed as:
\[ P = F \cdot v \]
where \( F \) is the force applied, and \( v \) is the velocity of the block.

### Step 2: Relate force to mass and acceleration
Using Newton's second law, we know:
\[ F = m \cdot a \]
where \( m \) is the mass of the block and \( a \) is its acceleration. The acceleration can be expressed as:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ F = m \cdot \frac{dv}{dt} \]

### Step 3: Substitute force into the power equation
Substituting \( F \) into the power equation gives:
\[ P = m \cdot \frac{dv}{dt} \cdot v \]

### Step 4: Rearranging the equation
We can rearrange this to isolate \( dt \):
\[ P \cdot dt = m \cdot v \cdot dv \]

### Step 5: Integrate both sides
Now we integrate both sides. The left side integrates with respect to time \( t \), and the right side integrates with respect to velocity \( v \):
\[ \int P \, dt = \int m v \, dv \]
This results in:
\[ Pt = \frac{m v^2}{2} \]
From this, we can express \( v^2 \) in terms of \( t \):
\[ v^2 = \frac{2Pt}{m} \]

### Step 6: Relate velocity to displacement
We know that velocity \( v \) is the rate of change of displacement \( x \):
\[ v = \frac{dx}{dt} \]
Substituting for \( v \) gives:
\[ \frac{dx}{dt} = \sqrt{\frac{2Pt}{m}} \]

### Step 7: Rearranging and integrating to find displacement
Rearranging gives:
\[ dx = \sqrt{\frac{2P}{m}} \cdot t^{1/2} dt \]
Integrating both sides:
\[ \int dx = \sqrt{\frac{2P}{m}} \int t^{1/2} dt \]
The integral of \( t^{1/2} \) is:
\[ x = \sqrt{\frac{2P}{m}} \cdot \left(\frac{t^{3/2}}{3/2}\right) + C \]
This simplifies to:
\[ x = \frac{2\sqrt{2P}}{3\sqrt{m}} t^{3/2} + C \]
Ignoring the constant of integration \( C \) for this relationship, we find:
\[ x \propto t^{3/2} \]

### Conclusion
Thus, the relation between displacement \( x \) and time \( t \) when a block is pulled with constant power is:
\[ x \propto t^{3/2} \]