To solve the problem step by step, we will break it down into two parts: (a) calculating the kinetic energy of the particle with respect to the passenger at the end of 8 seconds, and (b) calculating the kinetic energy of the particle for an observer on the ground.
### Part (a): Kinetic Energy with Respect to the Passenger
1. **Determine the acceleration of the train**:
The train has a constant acceleration \( a = 2.5 \, \text{m/s}^2 \).
2. **Calculate the velocity of the train after 8 seconds**:
\[
v_{\text{train}} = u + at = 0 + (2.5 \, \text{m/s}^2)(8 \, \text{s}) = 20 \, \text{m/s}
\]
3. **Determine the forces acting on the particle**:
- The horizontal force applied to the particle is \( F = 13 \, \text{N} \).
- The mass of the particle is \( m = 2 \, \text{kg} \).
- The normal force \( N \) acting on the particle is equal to its weight, \( N = mg = 2 \times 10 = 20 \, \text{N} \).
- The frictional force \( f_k \) (kinetic friction) is given by:
\[
f_k = \mu_k N = 0.5 \times 20 = 10 \, \text{N}
\]
4. **Calculate the net force acting on the particle**:
\[
F_{\text{net}} = F - f_k = 13 \, \text{N} - 10 \, \text{N} = 3 \, \text{N}
\]
5. **Calculate the acceleration of the particle**:
\[
a_{\text{particle}} = \frac{F_{\text{net}}}{m} = \frac{3 \, \text{N}}{2 \, \text{kg}} = 1.5 \, \text{m/s}^2
\]
6. **Determine the velocity of the particle in the direction perpendicular to the train after 2 seconds** (force is applied for 2 seconds):
\[
v_{\text{particle}} = u + at = 0 + (1.5 \, \text{m/s}^2)(2 \, \text{s}) = 3 \, \text{m/s}
\]
7. **Calculate the kinetic energy of the particle with respect to the passenger**:
The passenger is moving with the train, so the particle's velocity relative to the passenger is only in the perpendicular direction.
\[
KE_{\text{passenger}} = \frac{1}{2} m v^2 = \frac{1}{2} (2 \, \text{kg}) (3 \, \text{m/s})^2 = \frac{1}{2} (2)(9) = 9 \, \text{J}
\]
### Part (b): Kinetic Energy with Respect to an Observer on the Ground
1. **Calculate the velocity of the particle in the direction of the train after 8 seconds**:
The train's velocity is \( 20 \, \text{m/s} \) (calculated earlier), and the particle has an acceleration of \( 1.5 \, \text{m/s}^2 \) for 2 seconds:
\[
v_{\text{particle, horizontal}} = 0 + (2.5 \, \text{m/s}^2)(6 \, \text{s}) + (1.5 \, \text{m/s}^2)(2 \, \text{s}) = 15 + 3 = 18 \, \text{m/s}
\]
2. **Determine the total velocity of the particle**:
The particle's velocities in both directions:
- Horizontal (direction of train): \( 20 \, \text{m/s} \)
- Perpendicular (after 8 seconds): \( 3 \, \text{m/s} \)
Using Pythagorean theorem:
\[
v_{\text{total}} = \sqrt{(20 \, \text{m/s})^2 + (3 \, \text{m/s})^2} = \sqrt{400 + 9} = \sqrt{409} \, \text{m/s}
\]
3. **Calculate the kinetic energy of the particle with respect to the observer on the ground**:
\[
KE_{\text{ground}} = \frac{1}{2} m v^2 = \frac{1}{2} (2 \, \text{kg}) (409) = 409 \, \text{J}
\]
### Final Answers:
(a) Kinetic Energy with respect to the passenger: **9 J**
(b) Kinetic Energy with respect to the observer on the ground: **409 J**
