At what height above the ground must a mass of 5 kg be to have its PE equal in value to the KE possessed by it when it moves with a velocity of 10 m/s?`(g = 10 m/s^(2))`

Options

A. `1m`
B. `5m`
C. `10m`
D. `50m`

Explanation

To find the height above the ground at which the potential energy (PE) of a mass of 5 kg is equal to its kinetic energy (KE) when it moves with a velocity of 10 m/s, we can follow these steps:

### Step 1: Calculate the Kinetic Energy (KE)
The formula for kinetic energy is given by:
\[ KE = \frac{1}{2} mv^2 \]
Where:
- \( m = 5 \, \text{kg} \) (mass)
- \( v = 10 \, \text{m/s} \) (velocity)

Substituting the values:
\[ KE = \frac{1}{2} \times 5 \times (10)^2 \]
\[ KE = \frac{1}{2} \times 5 \times 100 \]
\[ KE = \frac{500}{2} = 250 \, \text{J} \]

### Step 2: Set the Potential Energy (PE) Equal to the Kinetic Energy (KE)
The formula for potential energy is given by:
\[ PE = mgh \]
Where:
- \( m = 5 \, \text{kg} \) (mass)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h \) is the height we need to find.

Since we want the potential energy to equal the kinetic energy:
\[ mgh = KE \]
Substituting the known values:
\[ 5 \times 10 \times h = 250 \]

### Step 3: Solve for Height (h)
Now, we can solve for \( h \):
\[ 50h = 250 \]
Dividing both sides by 50:
\[ h = \frac{250}{50} = 5 \, \text{m} \]

### Conclusion
The height above the ground at which the potential energy of the mass is equal to its kinetic energy is:
\[ \boxed{5 \, \text{m}} \]

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