An engine is working at a constant power draws a load of mass m against a resistance r. Find maximum speed of load and time taken to attain half this speed.

Options

A. (a)`t=(Pm)/(8r^(2))`
B. (b)`t=(Pm)/(8r)`
C. (c)`t=(Pm)/(r^(2))`
D. (d)`t=(Pm)/(9r^(2))`

Answer

To solve the problem of an engine working at a constant power that draws a load of mass \( m \) against a resistance \( r \), we need to find the maximum speed of the load and the time taken to attain half of this speed.

### Step-by-Step Solution:

1. **Understanding Power and Force Relationship**:
The power \( P \) of the engine is given by the formula:
\[ P = F \cdot v \]
where \( F \) is the force exerted by the engine and \( v \) is the velocity of the load. In this case, the force \( F \) is equal to the resistance \( r \) that the engine has to overcome.

2. **Setting Up the Equation**:
Since the engine is working against the resistance, we can write:
\[ P = r \cdot v \]
Rearranging this gives us:
\[ v = \frac{P}{r} \]
This equation shows that the maximum speed \( v_{\text{max}} \) of the load is directly proportional to the power and inversely proportional to the resistance.

3. **Finding Maximum Speed**:
From the equation derived, we find the maximum speed:
\[ v_{\text{max}} = \frac{P}{r} \]

4. **Finding Time to Attain Half the Speed**:
We need to find the time taken to reach half of the maximum speed, which is:
\[ v = \frac{v_{\text{max}}}{2} = \frac{P}{2r} \]

5. **Using the Relationship Between Force, Mass, and Acceleration**:
The force can also be expressed using Newton's second law:
\[ F = m \cdot a \]
where \( a \) is the acceleration. Since \( a = \frac{dv}{dt} \), we can write:
\[ r = m \cdot \frac{dv}{dt} \]

6. **Rearranging and Integrating**:
Rearranging gives:
\[ m \cdot dv = r \cdot dt \]
Integrating both sides, we set the limits for \( v \) from \( 0 \) to \( \frac{P}{2r} \) and for \( t \) from \( 0 \) to \( t \):
\[ \int_0^{\frac{P}{2r}} m \, dv = \int_0^t r \, dt \]
This results in:
\[ m \cdot \left(\frac{P}{2r} - 0\right) = r \cdot t \]
Simplifying gives:
\[ \frac{mP}{2r} = rt \]
Therefore, solving for \( t \):
\[ t = \frac{mP}{2r^2} \]

### Final Answers:
- **Maximum Speed**: \( v_{\text{max}} = \frac{P}{r} \)
- **Time to attain half the speed**: \( t = \frac{mP}{2r^2} \)