To solve the problem of determining the speed of the object at the bottom of the incline, we can apply the principles of energy conservation, factoring in work done by friction. Here's a breakdown of the steps:

Given:

• Mass of the object, $m = 2.00 \, \text{kg}$
• Initial velocity at the top, $v_0 = 1.00 \, \text{m/s}$
• Length of the incline, $d = 12.0 \, \text{m}$
• Height of the incline, $h = 5.00 \, \text{m}$
• Coefficient of kinetic friction, $\mu_k = 0.3$
• Gravitational acceleration, $g = 9.81 \, \text{m/s}^2$

Step 1: Potential energy at the top of the incline

The potential energy (PE) at the top is given by: $\text{PE}_{\text{top}} = mgh$ $\text{PE}_{\text{top}} = 2.00 \times 9.81 \times 5.00 = 98.1 \, \text{J}$

Step 2: Kinetic energy at the top of the incline

The initial kinetic energy (KE) is given by: $\text{KE}_{\text{top}} = \frac{1}{2}mv_0^2$ $\text{KE}_{\text{top}} = \frac{1}{2} \times 2.00 \times (1.00)^2 = 1.00 \, \text{J}$

Step 3: Work done by friction

The work done by friction is: $W_{\text{friction}} = - \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d$ We first need to find the angle of the incline, $\theta$, from the triangle formed by the ramp: $\sin(\theta) = \frac{h}{d} = \frac{5.00}{12.0}$ $\theta = \sin^{-1}\left( \frac{5.00}{12.0} \right) \approx 24.6^\circ$ Thus, $\cos(\theta) = \cos(24.6^\circ) \approx 0.906$.

Now, the work done by friction: $W_{\text{friction}} = - 0.3 \times 2.00 \times 9.81 \times 0.906 \times 12.0 \approx -63.94 \, \text{J}$

Step 4: Total mechanical energy at the bottom

At the bottom of the incline, the potential energy is zero. The total mechanical energy is the sum of the initial kinetic and potential energy, minus the work done by friction: $\text{Total Energy}_{\text{bottom}} = (\text{PE}_{\text{top}} + \text{KE}_{\text{top}}) + W_{\text{friction}}$ $\text{Total Energy}_{\text{bottom}} = (98.1 + 1.00) + (-63.94) = 35.16 \, \text{J}$

Step 5: Solve for the final velocity

The kinetic energy at the bottom is: $\text{KE}_{\text{bottom}} = \frac{1}{2}mv_{\text{bottom}}^2 = 35.16 \, \text{J}$ Solving for $v_{\text{bottom}}$: $v_{\text{bottom}} = \sqrt{\frac{2 \times 35.16}{2.00}} = \sqrt{35.16} \approx 5.93 \, \text{m/s}$

Final Answer:

The speed of the object at the bottom of the incline is approximately $5.93 \, \text{m/s}$.

Would you like more details on any of the steps, or do you have any questions?

Follow-up Questions:

1. How does the angle of incline affect the speed of the object at the bottom?
2. What would happen if the coefficient of friction were zero?
3. How can we calculate the time it takes for the object to reach the bottom?
4. How would the result change if the object started from rest?
5. What other forces might be considered in a real-world scenario (e.g., air resistance)?

Tip:

Always check the units and signs when calculating work done by non-conservative forces, such as friction. Negative work reduces the total mechanical energy.