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., taking into account sliding friction.

We are asked to find the linear velocity, angular velocity, and linear acceleration of a wheel subject to an applied force, rolling and sliding friction after 5 and 10 seconds.

To solve the problem, we'll use the following approach:

### Step 1: Define the forces and equations
- **Force applied to the center of the wheel:** \( F = 4 \, \text{N} \)
- **Mass of the wheel:** \( m = 3 \, \text{kg} \)
- **Radius of the wheel:** \( R = 0.3 \, \text{m} \)
- **Initial velocity:** \( V_0 = 2 \, \text{m/s} \)
- **Rolling friction coefficient:** \( K_1 = 0.016 \)
- **Sliding friction coefficient:** \( K_2 = 0.6 \)
- **Acceleration due to gravity:** \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Define the forces acting on the wheel
1. **Applied force:** \( F_{\text{applied}} = 4 \, \text{N} \).
2. **Rolling friction force:** The rolling friction acts in the opposite direction to the motion and is given by:
   \[
   F_{\text{rolling}} = K_1 \cdot N = K_1 \cdot m \cdot g
   \]
   where \( N = m \cdot g \) is the normal force.
3. **Sliding friction force:** If there is slipping (sliding) between the wheel and the surface, the sliding friction is given by:
   \[
   F_{\text{sliding}} = K_2 \cdot N = K_2 \cdot m \cdot g
   \]

### Step 3: Compute the net force
The net force acting on the wheel is the applied force minus the frictional forces. Thus, for sliding friction:
\[
F_{\text{net}} = F_{\text{applied}} - F_{\text{sliding}}
\]

Substitute the values:
\[
F_{\text{sliding}} = 0.6 \cdot 3 \cdot 9.8 = 17.64 \, \text{N}
\]
Thus,
\[
F_{\text{net}} = 4 - 17.64 = -13.64 \, \text{N}
\]
Since \( F_{\text{net}} \) is negative, the wheel is decelerating due to sliding friction.

### Step 4: Linear acceleration
The linear acceleration \( a \) is given by Newton's second law:
\[
a = \frac{F_{\text{net}}}{m} = \frac{-13.64}{3} = -4.55 \, \text{m/s}^2
\]

### Step 5: Angular acceleration
For pure rolling, the angular acceleration \( \alpha \) is related to the linear acceleration by:
\[
a = \alpha \cdot R
\]
Therefore,
\[
\alpha = \frac{a}{R} = \frac{-4.55}{0.3} = -15.17 \, \text{rad/s}^2
\]

### Step 6: Linear velocity after time \( t \)
The linear velocity after time \( t \) is given by:
\[
V(t) = V_0 + a \cdot t
\]
After 5 seconds:
\[
V(5) = 2 + (-4.55) \cdot 5 = 2 - 22.75 = -20.75 \, \text{m/s}
\]
After 10 seconds:
\[
V(10) = 2 + (-4.55) \cdot 10 = 2 - 45.5 = -43.5 \, \text{m/s}
\]

### Step 7: Angular velocity after time \( t \)
The angular velocity after time \( t \) is given by:
\[
\omega(t) = \omega_0 + \alpha \cdot t
\]
The initial angular velocity for pure rolling is related to the initial linear velocity by:
\[
\omega_0 = \frac{V_0}{R} = \frac{2}{0.3} = 6.67 \, \text{rad/s}
\]
After 5 seconds:
\[
\omega(5) = 6.67 + (-15.17) \cdot 5 = 6.67 - 75.85 = -69.18 \, \text{rad/s}
\]
After 10 seconds:
\[
\omega(10) = 6.67 + (-15.17) \cdot 10 = 6.67 - 151.7 = -145.03 \, \text{rad/s}
\]

### Step 8: Summary of Results
- **At \( t = 5 \, \text{s} \):**
  - Linear velocity: \( V(5) = -20.75 \, \text{m/s} \)
  - Angular velocity: \( \omega(5) = -69.18 \, \text{rad/s} \)
  - Linear acceleration: \( a = -4.55 \, \text{m/s}^2 \)

- **At \( t = 10 \, \text{s} \):**
  - Linear velocity: \( V(10) = -43.5 \, \text{m/s} \)
  - Angular velocity: \( \omega(10) = -145.03 \, \text{rad/s} \)
  - Linear acceleration: \( a = -4.55 \, \text{m/s}^2 \)

### Do you want further clarification or details on any step?

Here are five related questions for deeper exploration:
1. How would the results change if the rolling friction coefficient was larger?
2. What happens to the motion if there was no sliding friction, only rolling?
3. How would the angular momentum change with time for the wheel?
4. How does the presence of both rolling and sliding friction affect energy dissipation?
5. How would an increase in the applied force affect the wheel's motion?

**Tip:** For rotational motion, always relate linear and angular quantities using the radius \( R \), especially when considering rolling without slipping.

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., taking into account sliding friction.

Solve the problem of determining the linear and angular velocity, as well as acceleration, of a 3 kg wheel rolling on a flat surface under the influence of a 4 N force, considering both sliding and rolling friction. This solution involves kinematics, Newton's laws, and rotational dynamics, suitable for Grades 11-12 or undergraduate physics students.

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