Only one force is acting on a 2.8 kg particle-like object whose position is given by `x= 4.0t-5.0t^(2)+2.0t^(3)`, with `x` in meters and t in seconds. What is the work done by the force from `t=0` to `t= 6.0` s ?

Answer

To solve the problem, we need to find the work done by the force acting on a particle-like object from \( t = 0 \) to \( t = 6 \) seconds. We will use the work-energy theorem, which states that the work done by the total force is equal to the change in kinetic energy of the object.

### Step-by-Step Solution:

1. **Identify the Position Function**:
The position of the object is given by:
\[ x(t) = 4.0t - 5.0t^2 + 2.0t^3 \]

2. **Differentiate to Find Velocity**:
To find the velocity, we differentiate the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(4.0t - 5.0t^2 + 2.0t^3) \]
\[ v(t) = 4.0 - 10.0t + 6.0t^2 \]

3. **Calculate Initial Velocity \( v(0) \)**:
Substitute \( t = 0 \) into the velocity equation:
\[ v(0) = 4.0 - 10.0(0) + 6.0(0)^2 = 4.0 \, \text{m/s} \]

4. **Calculate Final Velocity \( v(6) \)**:
Substitute \( t = 6 \) into the velocity equation:
\[ v(6) = 4.0 - 10.0(6) + 6.0(6)^2 \]
\[ v(6) = 4.0 - 60.0 + 216.0 = 160.0 \, \text{m/s} \]

5. **Calculate Initial Kinetic Energy \( KE_i \)**:
The initial kinetic energy is given by:
\[ KE_i = \frac{1}{2}mv_i^2 \]
Where \( m = 2.8 \, \text{kg} \) and \( v_i = 4.0 \, \text{m/s} \):
\[ KE_i = \frac{1}{2}(2.8)(4.0)^2 = \frac{1}{2}(2.8)(16) = 22.4 \, \text{J} \]

6. **Calculate Final Kinetic Energy \( KE_f \)**:
The final kinetic energy is given by:
\[ KE_f = \frac{1}{2}mv_f^2 \]
Where \( v_f = 160.0 \, \text{m/s} \):
\[ KE_f = \frac{1}{2}(2.8)(160.0)^2 = \frac{1}{2}(2.8)(25600) = 35840 \, \text{J} \]

7. **Calculate Work Done \( W \)**:
The work done by the force is the change in kinetic energy:
\[ W = KE_f - KE_i = 35840 - 22.4 = 35817.6 \, \text{J} \]

8. **Convert to Kilojoules**:
\[ W = \frac{35817.6}{1000} = 35.8176 \, \text{kJ} \]

### Final Answer:
The work done by the force from \( t = 0 \) to \( t = 6 \) seconds is approximately:
\[ W \approx 35.82 \, \text{kJ} \]