A wheel with mass m = 3 kg, radius R = 0.3 m rolls on a flat surface under the influence of a force F = 4 N applied to the center of the wheel and directed parallel to the plane. Initial speed V0=2m/s.  Rolling friction coefficient K1=0,016.  Sliding friction coefficients K2=0.6.  Find the  linear, angular velocity and acceleration of the wheel after 5 s, 10 s. 

To solve this problem, we need to address both the linear and angular aspects of the motion of the rolling wheel. Let's break it down step by step.

### Given:
- Mass of the wheel: \( m = 3 \, \text{kg} \)
- Radius of the wheel: \( R = 0.3 \, \text{m} \)
- Applied force at the center: \( F = 4 \, \text{N} \)
- Initial velocity: \( V_0 = 2 \, \text{m/s} \)
- Rolling friction coefficient: \( K_1 = 0.016 \)
- Sliding friction coefficient: \( K_2 = 0.6 \)
- Gravitational acceleration: \( g = 9.81 \, \text{m/s}^2 \)

### Forces and Torques:
1. **Linear motion**: The net force affecting the linear acceleration comes from both the applied force \( F \) and the opposing rolling friction force \( F_{\text{fric}} \).
   \[
   F_{\text{fric}} = K_1 \cdot m \cdot g
   \]
   Therefore, the net force on the wheel is:
   \[
   F_{\text{net}} = F - F_{\text{fric}} = 4 - (0.016 \cdot 3 \cdot 9.81)
   \]
   
   Using Newton's second law, the linear acceleration \( a \) can be found as:
   \[
   a = \frac{F_{\text{net}}}{m}
   \]

2. **Angular motion**: The torque causing angular acceleration comes from the rolling friction at the point of contact between the wheel and the surface. The rolling friction produces a torque:
   \[
   \tau = F_{\text{fric}} \cdot R
   \]
   The moment of inertia \( I \) for a solid wheel is:
   \[
   I = \frac{1}{2} m R^2
   \]
   Using the relation between torque and angular acceleration \( \alpha \):
   \[
   \alpha = \frac{\tau}{I} = \frac{F_{\text{fric}} \cdot R}{\frac{1}{2} m R^2} = \frac{2 F_{\text{fric}}}{m R}
   \]

3. **Linear and angular velocity after time \( t \)**:
   - Linear velocity after time \( t \):
     \[
     V(t) = V_0 + a \cdot t
     \]
   - Angular velocity after time \( t \):
     \[
     \omega(t) = \omega_0 + \alpha \cdot t
     \]
     where the initial angular velocity \( \omega_0 \) is related to the initial linear velocity by:
     \[
     \omega_0 = \frac{V_0}{R}
     \]

### Step 1: Calculate rolling friction force and net force
The rolling friction force is:
\[
F_{\text{fric}} = 0.016 \cdot 3 \cdot 9.81 = 0.471 \, \text{N}
\]
The net force is:
\[
F_{\text{net}} = 4 - 0.471 = 3.529 \, \text{N}
\]

### Step 2: Calculate linear acceleration
Using \( a = \frac{F_{\text{net}}}{m} \):
\[
a = \frac{3.529}{3} = 1.176 \, \text{m/s}^2
\]

### Step 3: Calculate angular acceleration
The torque due to friction is:
\[
\tau = 0.471 \cdot 0.3 = 0.1413 \, \text{Nm}
\]
The angular acceleration is:
\[
\alpha = \frac{2 \cdot 0.471}{3 \cdot 0.3} = 1.047 \, \text{rad/s}^2
\]

### Step 4: Calculate velocities after 5 s and 10 s
1. **Linear velocity**:
   After 5 seconds:
   \[
   V(5) = 2 + 1.176 \cdot 5 = 7.88 \, \text{m/s}
   \]
   After 10 seconds:
   \[
   V(10) = 2 + 1.176 \cdot 10 = 13.76 \, \text{m/s}
   \]

2. **Angular velocity**:
   The initial angular velocity is:
   \[
   \omega_0 = \frac{2}{0.3} = 6.67 \, \text{rad/s}
   \]
   After 5 seconds:
   \[
   \omega(5) = 6.67 + 1.047 \cdot 5 = 11.91 \, \text{rad/s}
   \]
   After 10 seconds:
   \[
   \omega(10) = 6.67 + 1.047 \cdot 10 = 17.14 \, \text{rad/s}
   \]

### Final Answers:
- After 5 seconds:
  - Linear velocity: \( V(5) = 7.88 \, \text{m/s} \)
  - Angular velocity: \( \omega(5) = 11.91 \, \text{rad/s} \)

- After 10 seconds:
  - Linear velocity: \( V(10) = 13.76 \, \text{m/s} \)
  - Angular velocity: \( \omega(10) = 17.14 \, \text{rad/s} \)

Would you like further clarification on any part of the process?

### Related Questions:
1. What if the initial velocity were zero? How would the results change?
2. How would increasing the mass of the wheel affect the linear and angular accelerations?
3. What if the sliding friction coefficient \( K_2 \) played a role? How would that alter the calculations?
4. How can the moment of inertia change if the wheel is hollow instead of solid?
5. How does the torque from friction affect rotational kinetic energy?

### Tip:
Always consider the moment of inertia when dealing with rolling objects—it influences how the mass distribution affects rotational motion!

A wheel with mass m = 3 kg, radius R = 0.3 m rolls on a flat surface under the influence of a force F = 4 N applied to the center of the wheel and directed parallel to the plane. Initial speed V0=2m/s. Rolling friction coefficient K1=0.016. Sliding friction coefficient K2=0.6. Find the linear, angular velocity and acceleration of the wheel after 5 s, 10 s.

Solve for the linear and angular velocity of a wheel rolling on a flat surface under a force of 4 N. The wheel has a mass of 3 kg, a radius of 0.3 m, and experiences rolling friction with a coefficient of 0.016. Learn how to calculate acceleration and velocity after 5 s and 10 s using physics concepts such as torque, moment of inertia, and Newton's laws.

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