A block of ice with mass 2.00 kg slides 0.750 m down an inclined
plane that slopes downward at an angle of below the horizontal. If the
block of ice starts from rest, what is its final speed? You can ignore
friction.

To solve this problem, we'll apply the principle of **conservation of mechanical energy**. Since friction is neglected, the mechanical energy (sum of potential and kinetic energy) is conserved.

### Step-by-Step Solution:

1. **Initial Energy**:
   The block of ice starts from rest, so its initial kinetic energy is zero. It also has potential energy due to its position on the inclined plane.

   Initial mechanical energy, \(E_{\text{initial}}\), is just the potential energy:
   \[
   E_{\text{initial}} = U_i = mgh
   \]
   where:
   - \(m = 2.00 \, \text{kg}\) (mass of the ice block),
   - \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity),
   - \(h\) is the vertical height change.

2. **Height Calculation**:
   The block slides down a distance of 0.750 m along the inclined plane. The vertical height change is related to the distance traveled along the plane and the angle of the incline, \(\theta\), by:
   \[
   h = d \sin \theta
   \]
   where \(d = 0.750 \, \text{m}\) is the distance along the plane. We will leave \(\theta\) as the angle for now.

3. **Final Energy**:
   At the bottom, all the potential energy is converted into kinetic energy since the block is now at a lower height. The final mechanical energy is:
   \[
   E_{\text{final}} = K_f = \frac{1}{2} m v_f^2
   \]
   where \(v_f\) is the final velocity of the block.

4. **Conservation of Energy**:
   Since there is no friction, the total mechanical energy is conserved. Therefore:
   \[
   E_{\text{initial}} = E_{\text{final}}
   \]
   \[
   mgh = \frac{1}{2} m v_f^2
   \]
   The mass cancels out from both sides:
   \[
   gh = \frac{1}{2} v_f^2
   \]
   Solving for \(v_f\):
   \[
   v_f = \sqrt{2gh}
   \]
   Substituting \(h = d \sin \theta\):
   \[
   v_f = \sqrt{2g d \sin \theta}
   \]
   Now, substituting the known values:
   \[
   v_f = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 0.750 \, \text{m} \times \sin \theta}
   \]
   To get the final speed, you'll need the value of \(\theta\) (the angle of the incline). If you provide this, I can calculate the final speed precisely.

Would you like to give the value of \(\theta\) or have any clarifications? 

### 5 Related Questions:
1. How does friction affect the conservation of mechanical energy in inclined plane problems?
2. How would the final speed change if the incline were steeper?
3. What is the role of the angle in determining the height in inclined plane problems?
4. How can we determine the time it takes for the block to reach the bottom of the incline?
5. How does this problem relate to the concept of work and energy?

**Tip:** Always ensure to work with consistent units (e.g., meters, seconds) when using energy equations to avoid errors.

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of below the horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction.

Learn how to calculate the final speed of a 2.00 kg block of ice sliding 0.750 m down an inclined plane, using conservation of energy and trigonometric principles. This step-by-step guide explains the key formulas and concepts, such as gravitational potential energy, kinematics, and the role of incline angles. Suitable for Grades 11-12 students.

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