The total mechanical energy of an object with mass m at height h in a closed system can be written as ME=1/2mv2+mgh.  Solve for v, the velocity of the object, in terms of the given quantities.

To solve for the velocity \( v \) in terms of the given quantities in the equation for total mechanical energy:

\[
ME = \frac{1}{2}mv^2 + mgh
\]

we will follow these steps:

### Step 1: Subtract the potential energy term (\( mgh \)) from both sides.
\[
ME - mgh = \frac{1}{2}mv^2
\]

### Step 2: Multiply both sides by 2 to eliminate the fraction.
\[
2(ME - mgh) = mv^2
\]

### Step 3: Divide both sides by the mass (\( m \)) to isolate \( v^2 \).
\[
\frac{2(ME - mgh)}{m} = v^2
\]

### Step 4: Take the square root of both sides to solve for \( v \).
\[
v = \sqrt{\frac{2(ME - mgh)}{m}}
\]

Thus, the velocity \( v \) of the object in terms of the given quantities is:

\[
v = \sqrt{\frac{2(ME - mgh)}{m}}
\]

### Do you have any questions or would you like further details?

Here are 5 related questions to expand on this concept:
1. What happens to the velocity if the mechanical energy is entirely kinetic (i.e., \( h = 0 \))?
2. How does increasing the height \( h \) affect the velocity \( v \)?
3. Can the velocity be zero if there is mechanical energy in the system? Under what conditions?
4. How does this equation change if friction or air resistance is introduced to the system?
5. What is the physical interpretation of mechanical energy in a non-closed system?

**Tip:** Mechanical energy in a closed system remains constant, meaning any increase in kinetic energy results in a decrease in potential energy, and vice versa.

The total mechanical energy of an object with mass m at height h in a closed system can be written as ME=1/2mv^2+mgh. Solve for v, the velocity of the object, in terms of the given quantities.

Learn how to solve for the velocity of an object in a closed system, given its mechanical energy, mass, and height. This problem covers algebraic manipulation of energy equations, including kinetic and potential energy. Suitable for students in Grades 10-12.

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