### Problem StatementA block is initially held at rest at the bottom of a very long slide that is angled at 40 degrees with respect to the horizontal. There is no friction between the block and the slide.The block is then 'kicked' so that it begins up the slide with an initial speed equal to 3 m/s.Using only the principles of conservation of energy and geometry/trigonometry (do not use Free body diagrams or kinematic equations), determine the distance \( d \) that the block travels up the slide before it begins to come back down again.### Diagram ExplanationThe diagram shows an inclined plane (slide) with an angle \(\theta = 40^\circ\) from the horizontal. A block is placed at the bottom of this slide. An arrow indicates the direction of the block's initial velocity up the slide. The slide length along which the block travels before stopping is labeled as \(d\).### Solution Approach1. **Conservation of Energy**: - **Initial Energy**: The block starts with kinetic energy due to its speed. - **Final Energy**: At its highest point on the slide, the block's speed is 0, and all kinetic energy is converted into potential energy.2. **Energy Equations**: - Initial kinetic energy: \( \frac{1}{2} m v^2 \) - Final potential energy: \( mgh \)3. **Geometry**: - \( h = d \sin(\theta) \)4. **Set Energy Equal**: Solve for \(d\) using conservation of energy: \[ \frac{1}{2} m v^2 = mgd \sin(\theta) \] Simplify and solve for \(d\): \[ d = \frac{v^2}{2g \sin(\theta)} \]By substituting the known values: - \( v = 3 \, \text{m/s} \) - \(\theta = 40^\circ\) - \( g = 9.8 \, \text{m/s}^2 \)Calculate \( d \).

Name: ### Problem StatementA block is initially held at rest at the bottom
Uploaded: 2024-03-20T06:57:57.000Z
Channel: Bartleby
Description: ### Problem StatementA block is initially held at rest at the bottom of a very long slide that is angled at 40 degrees with respect to the horizontal. There is no friction between the block and the slide.The block is then 'kicked' so that it begins