Questions: As the block slides down the ramp, what happens to its kinetic energy K, potential energy U, and total mechanical energy E? K decreases, U increases E stays the same K decreases, U increases E increases K increases; U increases; E increases increases E decreases E stays the same

As the block slides down the ramp, what happens to its kinetic energy K, potential energy U, and total mechanical energy E?
K decreases, U increases E stays the same
K decreases, U increases E increases
K increases; U increases; E increases
increases E decreases E stays the same
Transcript text: As the block slides down the ramp, what happens to its kinetic energy $K$, potential energy $U$, and total mechanical energy E? $K$ decreases,$U$ increases $E$ stays the same $K$ decreases,$U$ increases $E$ increases $K$ increases; $U$ increases; $E$ increases increases $E$ decreases $E$ stays the same
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Solution

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Solution Steps

Step 1: Understanding the Problem

We need to analyze the changes in kinetic energy \( K \), potential energy \( U \), and total mechanical energy \( E \) of a block sliding down a ramp. The options provided describe different scenarios of how these energies change.

Step 2: Analyzing Energy Changes
  • Kinetic Energy (\( K \)): As the block slides down the ramp, it accelerates due to gravity, which means its speed increases. Therefore, the kinetic energy \( K \) increases.
  • Potential Energy (\( U \)): The potential energy is given by \( U = mgh \), where \( h \) is the height. As the block descends, \( h \) decreases, so the potential energy \( U \) decreases.
  • Total Mechanical Energy (\( E \)): In the absence of non-conservative forces like friction, the total mechanical energy \( E = K + U \) remains constant.
Step 3: Matching with Given Options

Based on the analysis:

  • \( K \) increases
  • \( U \) decreases
  • \( E \) stays the same

None of the provided options exactly match this scenario. However, the closest option that aligns with the conservation of energy principle is that the total mechanical energy \( E \) stays the same.

Final Answer

\[ \boxed{K \text{ increases, } U \text{ decreases, } E \text{ stays the same}} \]

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