Power applied to a particle varies with time as `P=(4t^(3)-5t+2)` watt, where t is in second. Find the change its K.F. between time t = 2 and t = 4 sec.

Options

A. 212 J
B. 213 J
C. 214 J
D. 215 J

Answer

To find the change in kinetic energy (K.E.) of a particle when the power applied to it varies with time as \( P = 4t^3 - 5t + 2 \) watts, we can follow these steps:

### Step 1: Understand the relationship between power and work done
Power is defined as the rate of work done, which can be expressed mathematically as:
\[ P = \frac{dW}{dt} \]
From this, we can derive the infinitesimal work done \( dW \):
\[ dW = P \, dt \]

### Step 2: Substitute the expression for power into the work done equation
Substituting the given expression for power into the work done equation, we have:
\[ dW = (4t^3 - 5t + 2) \, dt \]

### Step 3: Integrate to find the total work done
To find the total work done \( W \) between the time intervals \( t = 2 \) seconds and \( t = 4 \) seconds, we need to integrate \( dW \):
\[ W = \int_{2}^{4} (4t^3 - 5t + 2) \, dt \]

### Step 4: Perform the integration
Now we will integrate the function:
\[ W = \int (4t^3 - 5t + 2) \, dt = \left( \frac{4t^4}{4} - \frac{5t^2}{2} + 2t \right) + C \]
This simplifies to:
\[ W = t^4 - \frac{5t^2}{2} + 2t \]
Now, we will evaluate this from \( t = 2 \) to \( t = 4 \):
\[ W(4) = 4^4 - \frac{5 \cdot 4^2}{2} + 2 \cdot 4 \]
Calculating each term:
- \( 4^4 = 256 \)
- \( \frac{5 \cdot 16}{2} = 40 \)
- \( 2 \cdot 4 = 8 \)

Thus,
\[ W(4) = 256 - 40 + 8 = 224 \]

Now, for \( t = 2 \):
\[ W(2) = 2^4 - \frac{5 \cdot 2^2}{2} + 2 \cdot 2 \]
Calculating each term:
- \( 2^4 = 16 \)
- \( \frac{5 \cdot 4}{2} = 10 \)
- \( 2 \cdot 2 = 4 \)

Thus,
\[ W(2) = 16 - 10 + 4 = 10 \]

### Step 5: Calculate the change in work done
The change in work done \( \Delta W \) from \( t = 2 \) to \( t = 4 \) is:
\[ \Delta W = W(4) - W(2) = 224 - 10 = 214 \]

### Step 6: Relate work done to change in kinetic energy
According to the work-energy theorem, the work done on an object is equal to the change in kinetic energy:
\[ \Delta K.E. = \Delta W \]

Thus, the change in kinetic energy is:
\[ \Delta K.E. = 214 \, \text{J} \]

### Final Answer
The change in kinetic energy between time \( t = 2 \) seconds and \( t = 4 \) seconds is \( 214 \, \text{J} \).

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