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. 

Let's solve this problem step-by-step by addressing the forces and torques acting on the rolling wheel, using Newton's laws for both linear and rotational motion.

### Given Data:
- \( m = 3 \, \text{kg} \) (mass of the wheel)
- \( R = 0.3 \, \text{m} \) (radius of the wheel)
- \( F = 4 \, \text{N} \) (force applied at the center of the wheel)
- \( V_0 = 2 \, \text{m/s} \) (initial speed)
- Rolling friction coefficient \( K_1 = 0.016 \)
- Sliding friction coefficient \( K_2 = 0.6 \)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)

### 1. Linear Motion Analysis

The net force acting on the wheel is:
\[
F_{\text{net}} = F - f_{\text{rolling}}
\]
where \( f_{\text{rolling}} = K_1 \cdot mg \) is the rolling friction force. Substituting the given values:
\[
f_{\text{rolling}} = 0.016 \cdot 3 \cdot 9.81 = 0.471 \, \text{N}
\]
Thus, the net force:
\[
F_{\text{net}} = 4 - 0.471 = 3.529 \, \text{N}
\]

Now, the linear acceleration \( a_{\text{linear}} \) is given by Newton's second law:
\[
a_{\text{linear}} = \frac{F_{\text{net}}}{m} = \frac{3.529}{3} = 1.1763 \, \text{m/s}^2
\]

### 2. Angular Motion Analysis

The torque \( \tau \) acting on the wheel is produced by the rolling friction force at the point of contact with the ground. This is given by:
\[
\tau = f_{\text{rolling}} \cdot R
\]
Substituting the values:
\[
\tau = 0.471 \cdot 0.3 = 0.1413 \, \text{Nm}
\]

The moment of inertia of a solid disk (assuming the wheel is a disk) is:
\[
I = \frac{1}{2} m R^2 = \frac{1}{2} \cdot 3 \cdot (0.3)^2 = 0.135 \, \text{kg} \cdot \text{m}^2
\]

Now, the angular acceleration \( \alpha \) is given by:
\[
\alpha = \frac{\tau}{I} = \frac{0.1413}{0.135} = 1.046 \, \text{rad/s}^2
\]

### 3. Finding Velocity and Angular Velocity After 5 s and 10 s

#### After 5 seconds:
- The linear velocity after time \( t = 5 \, \text{s} \) is:
  \[
  V(t) = V_0 + a_{\text{linear}} \cdot t = 2 + 1.1763 \cdot 5 = 7.8815 \, \text{m/s}
  \]

- The angular velocity after time \( t = 5 \, \text{s} \) is:
  \[
  \omega(t) = \omega_0 + \alpha \cdot t
  \]
  The initial angular velocity \( \omega_0 \) is related to the initial linear velocity \( V_0 \):
  \[
  \omega_0 = \frac{V_0}{R} = \frac{2}{0.3} = 6.67 \, \text{rad/s}
  \]
  Thus, the angular velocity after 5 s is:
  \[
  \omega(t) = 6.67 + 1.046 \cdot 5 = 11.9 \, \text{rad/s}
  \]

#### After 10 seconds:
- The linear velocity after time \( t = 10 \, \text{s} \) is:
  \[
  V(t) = V_0 + a_{\text{linear}} \cdot t = 2 + 1.1763 \cdot 10 = 13.763 \, \text{m/s}
  \]

- The angular velocity after time \( t = 10 \, \text{s} \) is:
  \[
  \omega(t) = 6.67 + 1.046 \cdot 10 = 16.13 \, \text{rad/s}
  \]

### Summary of Results:
- **After 5 seconds**:
  - Linear velocity: \( V(5) = 7.88 \, \text{m/s} \)
  - Angular velocity: \( \omega(5) = 11.9 \, \text{rad/s} \)

- **After 10 seconds**:
  - Linear velocity: \( V(10) = 13.76 \, \text{m/s} \)
  - Angular velocity: \( \omega(10) = 16.13 \, \text{rad/s} \)

Let me know if you'd like further details on any part of the solution, or if you have additional questions.

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### Related Questions:
1. How does the rolling friction coefficient affect the wheel's motion over time?
2. What would happen if the applied force were increased to 6 N?
3. How would the moment of inertia change if the wheel had a different shape (e.g., a hoop instead of a disk)?
4. What are the effects of the sliding friction coefficient \( K_2 \) in case the wheel begins to slide?
5. How would the wheel’s velocity change if the initial velocity \( V_0 \) was 0?

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### Tip: 
Always ensure that the moment of inertia used in rotational motion problems corresponds to the correct shape of the object (disk, hoop, etc.). It significantly influences angular acceleration.

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.

Find the linear and angular velocity of a wheel with mass 3 kg and radius 0.3 m, rolling under a force of 4 N. This problem uses Newton's laws and rotational dynamics to solve for velocities and accelerations at 5 and 10 seconds. Suitable for Grades 10-12 physics students.

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