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. (a)212 J
B. (b)213 J
C. (c)214 J
D. (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 will follow these steps:

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

### Step 2: Integrate power to find work done
To find the work done over a time interval, we can integrate power with respect to time:
\[ W = \int P \, dt \]
Given \( P = 4t^3 - 5t + 2 \), we need to integrate this from \( t = 2 \) seconds to \( t = 4 \) seconds.

### Step 3: Set up the integral
We set up the integral as follows:
\[ W = \int_{2}^{4} (4t^3 - 5t + 2) \, dt \]

### Step 4: Perform the integration
Now we integrate each term:
\[ W = \left[ \frac{4}{4} t^4 - \frac{5}{2} t^2 + 2t \right]_{2}^{4} \]
This simplifies to:
\[ W = \left[ t^4 - \frac{5}{2} t^2 + 2t \right]_{2}^{4} \]

### Step 5: Evaluate the integral at the limits
Now we evaluate the expression at the upper limit \( t = 4 \) and the lower limit \( t = 2 \):
1. For \( t = 4 \):
\[ W(4) = 4^4 - \frac{5}{2} \cdot 4^2 + 2 \cdot 4 = 256 - \frac{5}{2} \cdot 16 + 8 \]
\[ = 256 - 40 + 8 = 224 \]

2. For \( t = 2 \):
\[ W(2) = 2^4 - \frac{5}{2} \cdot 2^2 + 2 \cdot 2 = 16 - \frac{5}{2} \cdot 4 + 4 \]
\[ = 16 - 10 + 4 = 10 \]

### Step 6: Calculate the work done
Now, we find the change in work done (which equals the change in kinetic energy):
\[ \Delta W = W(4) - W(2) = 224 - 10 = 214 \text{ joules} \]

### Final Answer
The change in kinetic energy between \( t = 2 \) seconds and \( t = 4 \) seconds is \( 214 \) joules.

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