To solve this problem, we need to analyze the energy changes of the block as it slides down the ramp with and without friction.

Given Data:

• Length of the friction section, $L = 0.75 \, \text{m}$
• Initial height, $h = 2.0 \, \text{m}$
• Angle of the ramp, $\theta = 29^\circ$
• Coefficient of kinetic friction, $\mu_k = 0.450$
• Speed at point A, $v_A = 8.6 \, \text{m/s}$

1. Calculate the gravitational potential energy at point A:

The potential energy at point A can be calculated using: $PE_A = mgh$ where $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$).

2. Calculate the kinetic energy at point A:

The kinetic energy at point A is given by: $KE_A = \frac{1}{2} mv_A^2$

3. Determine the work done against friction:

The work done against friction as the block moves down the length $L$ can be calculated as: $W_f = f_k L = \mu_k mg \cos(\theta) L$ where $f_k$ is the friction force, $mg \cos(\theta)$ is the normal force.

4. Set up the energy conservation equation:

As the block moves from point A to point B, the energy conservation equation (taking into account work done against friction) can be set up as follows: $PE_A + KE_A - W_f = KE_B$ If the block reaches point B, we can solve for $v_B$ (the speed at point B): $v_B = \sqrt{\frac{2}{m} \left( PE_A + KE_A - W_f \right)}$

5. If the block cannot reach point B:

If $v_B$ is less than or equal to zero, we calculate the maximum height the block can achieve by using energy conservation: $PE_A + KE_A - W_f = mgh_{max}$ Solving for $h_{max}$ will give us the maximum height above point A.

Calculations:

1. Potential Energy at A: $PE_A = mg(2.0) \quad (\text{we will use } m \text{ later})$

2. Kinetic Energy at A: $KE_A = \frac{1}{2} m (8.6)^2$

3. Work Done Against Friction: $W_f = \mu_k mg \cos(29^\circ) L = 0.450 \cdot mg \cdot \cos(29^\circ) \cdot 0.75$

4. Substituting and Solving: Now, substituting back into the energy conservation equation: $mg(2.0) + \frac{1}{2} m (8.6)^2 - 0.450 \cdot mg \cdot \cos(29^\circ) \cdot 0.75 = \frac{1}{2} m v_B^2$

Cancel $m$ from both sides and solve for $v_B$: $g(2.0) + \frac{1}{2} (8.6)^2 - 0.450 \cdot g \cdot \cos(29^\circ) \cdot 0.75 = \frac{1}{2} v_B^2$

After calculating the terms:

1. Calculate $g \cos(29^\circ)$ and other values:

   • $g \cos(29^\circ) \approx 9.81 \cdot 0.8746 \approx 8.59 \, \text{m/s}^2$

2. Final Calculation: Substituting known values back to get $v_B$.

Conclusion:

By computing all of the above, we can find the speed at point B or the maximum height above A if it doesn't make it to point B.

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