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=0,02.  Sliding friction coefficients K2=0.15.  Find the  linear, angular velocity and acceleration of the wheel after 5 s, 10 s. 

This problem involves a wheel rolling on a flat surface under the influence of both an applied force and friction. To solve this, we need to calculate the linear and angular velocities, as well as the linear and angular accelerations, at 5 seconds and 10 seconds. We'll account for both rolling friction and sliding friction in the process.

### Step 1: Force and Torque Analysis
1. **Given Data:**
   - Mass of the wheel: \( m = 3 \, \text{kg} \)
   - Radius of the wheel: \( R = 0.3 \, \text{m} \)
   - Applied force: \( F = 4 \, \text{N} \)
   - Initial speed: \( V_0 = 2 \, \text{m/s} \)
   - Rolling friction coefficient: \( K_1 = 0.02 \)
   - Sliding friction coefficient: \( K_2 = 0.15 \)
   - Moment of inertia for a solid disc (assuming the wheel is solid): 
   \[
   I = \frac{1}{2} m R^2
   \]

### Step 2: Calculating the Forces Involved
- **Rolling friction force**: 
\[
F_{\text{rolling}} = K_1 \cdot N = K_1 \cdot m \cdot g
\]
Where \( g = 9.8 \, \text{m/s}^2 \) is the gravitational acceleration.

- **Sliding friction force**: 
\[
F_{\text{sliding}} = K_2 \cdot N = K_2 \cdot m \cdot g
\]

The rolling and sliding friction forces oppose the motion of the wheel.

### Step 3: Newton's Second Law for Linear Motion
The net force acting on the wheel determines its linear acceleration:
\[
F_{\text{net}} = F - F_{\text{rolling}} - F_{\text{sliding}}
\]
Then, using Newton’s second law:
\[
a_{\text{linear}} = \frac{F_{\text{net}}}{m}
\]
The linear velocity after time \( t \) can be found using:
\[
v = V_0 + a_{\text{linear}} \cdot t
\]

### Step 4: Torque and Angular Acceleration
The applied force also generates a torque, which causes angular acceleration:
\[
\tau = F \cdot R
\]
Using the rotational equivalent of Newton’s second law:
\[
\tau = I \cdot \alpha
\]
Where \( I = \frac{1}{2} m R^2 \) is the moment of inertia and \( \alpha \) is the angular acceleration:
\[
\alpha = \frac{\tau}{I} = \frac{F \cdot R}{\frac{1}{2} m R^2} = \frac{2F}{mR}
\]
The angular velocity after time \( t \) is:
\[
\omega = \omega_0 + \alpha \cdot t
\]
Where \( \omega_0 = \frac{V_0}{R} \) is the initial angular velocity.

### Step 5: Calculate for 5 s and 10 s

Now, let's compute these values step by step.

#### 1. Friction forces:
\[
F_{\text{rolling}} = K_1 \cdot m \cdot g = 0.02 \cdot 3 \cdot 9.8 = 0.588 \, \text{N}
\]
\[
F_{\text{sliding}} = K_2 \cdot m \cdot g = 0.15 \cdot 3 \cdot 9.8 = 4.41 \, \text{N}
\]

#### 2. Net force:
\[
F_{\text{net}} = 4 - 0.588 - 4.41 = -0.998 \, \text{N}
\]
This negative value indicates that the friction forces are greater than the applied force, meaning the wheel will slow down due to friction.

#### 3. Linear acceleration:
\[
a_{\text{linear}} = \frac{F_{\text{net}}}{m} = \frac{-0.998}{3} = -0.333 \, \text{m/s}^2
\]

#### 4. Linear velocity at 5 s:
\[
v_5 = V_0 + a_{\text{linear}} \cdot 5 = 2 + (-0.333) \cdot 5 = 2 - 1.665 = 0.335 \, \text{m/s}
\]

#### 5. Linear velocity at 10 s:
\[
v_{10} = V_0 + a_{\text{linear}} \cdot 10 = 2 + (-0.333) \cdot 10 = 2 - 3.33 = -1.33 \, \text{m/s}
\]
(At 10 seconds, the wheel would stop and begin to slide in the opposite direction.)

#### 6. Angular acceleration:
\[
\alpha = \frac{2F}{mR} = \frac{2 \cdot 4}{3 \cdot 0.3} = 8.89 \, \text{rad/s}^2
\]

#### 7. Angular velocity at 5 s:
\[
\omega_0 = \frac{V_0}{R} = \frac{2}{0.3} = 6.67 \, \text{rad/s}
\]
\[
\omega_5 = \omega_0 + \alpha \cdot 5 = 6.67 + 8.89 \cdot 5 = 6.67 + 44.45 = 51.12 \, \text{rad/s}
\]

#### 8. Angular velocity at 10 s:
\[
\omega_{10} = \omega_0 + \alpha \cdot 10 = 6.67 + 8.89 \cdot 10 = 6.67 + 88.9 = 95.57 \, \text{rad/s}
\]

### Final Results:
1. At \( t = 5 \, \text{s} \):
   - Linear velocity: \( v_5 = 0.335 \, \text{m/s} \)
   - Angular velocity: \( \omega_5 = 51.12 \, \text{rad/s} \)
   - Linear acceleration: \( a_{\text{linear}} = -0.333 \, \text{m/s}^2 \)
   - Angular acceleration: \( \alpha = 8.89 \, \text{rad/s}^2 \)

2. At \( t = 10 \, \text{s} \):
   - Linear velocity: \( v_{10} = -1.33 \, \text{m/s} \)
   - Angular velocity: \( \omega_{10} = 95.57 \, \text{rad/s} \)
   - Linear acceleration: \( a_{\text{linear}} = -0.333 \, \text{m/s}^2 \)
   - Angular acceleration: \( \alpha = 8.89 \, \text{rad/s}^2 \)

Would you like further details or clarifications?

### Additional questions to explore:
1. How would different friction coefficients affect the linear velocity?
2. What would happen if the rolling friction was zero but sliding friction remained?
3. How does the moment of inertia affect the angular acceleration in different scenarios?
4. If the force applied was increased, how would the motion of the wheel change?
5. What if the initial velocity was zero—how long would it take for the wheel to start rolling?

#### Tip:
Always analyze both the forces and torques acting on a body when solving for its linear and angular 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=0,02. Sliding friction coefficient K2=0.15. Find the linear, angular velocity and acceleration of the wheel after 5 s, 10 s.

Solve for the linear and angular velocity of a 3 kg wheel with radius 0.3 m, subject to both rolling and sliding friction, under a 4 N force. Learn step-by-step how to compute these values after 5 and 10 seconds, using Newton's laws and rotational kinematics.

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