A body of mass 10 kg moves according to the relation `x = t^(2) + 2t^(3)`. The work done by the force in the first 2s is.

Options

A. 7840 j
B. 1960 j
C. 3920 j
D. 4840 j

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Determine the displacement function
The displacement of the body is given by the equation:
\[ x(t) = t^2 + 2t^3 \]

### Step 2: Calculate the velocity
Velocity is the derivative of displacement with respect to time. Therefore, we differentiate \( x(t) \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^2 + 2t^3) = 2t + 6t^2 \]

### Step 3: Calculate the initial and final velocities
- **Initial velocity** at \( t = 0 \):
\[ v(0) = 2(0) + 6(0)^2 = 0 \, \text{m/s} \]

- **Final velocity** at \( t = 2 \):
\[ v(2) = 2(2) + 6(2)^2 = 4 + 6 \times 4 = 4 + 24 = 28 \, \text{m/s} \]

### Step 4: Calculate the change in kinetic energy
The change in kinetic energy (\( \Delta KE \)) is given by:
\[ \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) \]
Where:
- \( m = 10 \, \text{kg} \)
- \( v_f = 28 \, \text{m/s} \)
- \( v_i = 0 \, \text{m/s} \)

Substituting the values:
\[ \Delta KE = \frac{1}{2} \times 10 \times (28^2 - 0^2) \]
\[ = 5 \times 784 = 3920 \, \text{J} \]

### Step 5: Conclusion
The work done by the force in the first 2 seconds is:
\[ \text{Work Done} = 3920 \, \text{J} \]