Question

A steel ball of mass 0.1 kg is loaded into a horizontal ‘spring gun’ with spring constant 1000 N/m. The compression of the spring is measured to be 0.01 m. When the spring is released, the spring force launches the ball. What is the horizontal acceleration of the ball upon release? g=10m/s^2

Answer 1

The horizontal acceleration of the ball upon release is
`\(100 \, \text{m/s}^2\).`

The force exerted by the spring (
`\(F_s\)`) can be calculated using Hooke's Law:

`\[ F_s = k \cdot \Delta x \]`

Where:

- k is the spring constant (1000 N/m),

-
`\( \Delta x \)` is the compression of the spring (0.01 m).

`\[ F_s = (1000 \, \text{N/m}) \cdot (0.01 \, \text{m}) = 10 \, \text{N} \]`

Now, using Newton's second law (
`\( F = m \cdot a \)`), we can find the acceleration (a) of the ball:

`\[ F = m \cdot a \]`

`\[ 10 \, \text{N} = (0.1 \, \text{kg}) \cdot a \]`

`\[ a = \frac{10 \, \text{N}}{0.1 \, \text{kg}} = 100 \, \text{m/s}^2 \]`

So, the horizontal acceleration of the ball upon release is
`\(100 \, \text{m/s}^2\).`