A ball of mass `m` is droppped from a height `h` on a platform fixed at the top of a vertical spring. The platform is depressed by a distance `x`. What is the spring constant `K`?

Options

A. `(2mg)/(x)`
B. `(2mgh)/(x^(2)`
C. `(2mg(h+x))/(x^(2)`
D. `(2mg(h+x))/(h^(2)`

Answer

To find the spring constant \( K \) when a ball of mass \( m \) is dropped from a height \( h \) onto a platform fixed at the top of a vertical spring, which is depressed by a distance \( x \), we can use the principle of conservation of energy.

### Step-by-Step Solution:

1. **Identify the energies involved**:
When the ball is dropped from height \( h \), it possesses gravitational potential energy given by:
\[ PE_{\text{initial}} = mgh \]
As the ball falls, this potential energy is converted into kinetic energy and then into the potential energy of the spring when the spring is compressed by distance \( x \).

2. **Calculate the total potential energy at maximum compression**:
At the point of maximum compression of the spring, the total distance fallen by the ball is \( h + x \). Therefore, the gravitational potential energy at this point is:
\[ PE_{\text{total}} = mg(h + x) \]

3. **Calculate the potential energy stored in the spring**:
The potential energy stored in the spring when it is compressed by distance \( x \) is given by:
\[ PE_{\text{spring}} = \frac{1}{2} K x^2 \]

4. **Apply the conservation of energy principle**:
According to the conservation of energy, the total gravitational potential energy lost by the ball will equal the potential energy stored in the spring:
\[ mg(h + x) = \frac{1}{2} K x^2 \]

5. **Rearranging the equation to solve for \( K \)**:
Rearranging the equation gives:
\[ K x^2 = 2mg(h + x) \]
Therefore, we can express \( K \) as:
\[ K = \frac{2mg(h + x)}{x^2} \]

### Final Result:
The spring constant \( K \) is given by:
\[ K = \frac{2mg(h + x)}{x^2} \]