Question

A springboard device consisting of a spring that is sandwiched between two wood boards (see figure) is used to launch a small cube straight up into the air. Each wood board has a length, width, and depth of L = 20.0 cm , W = 15.0 cm, D = 3.0 cm, and the spring has spring constant of 800 N/m. The cube has a mass of 0.150 kg and each side of the cube has a length of s = 6.00 cm. The cube is pressed downward until the solid pressure between the cube and the upper board is 24.6 kPa, and then it is released. Neglecting air resistance, calculate the maximum height gain (relative to it height when it is first released) obtained by the cube.

Answer 1

The maximum height gained by the cube is approximately
`\(0.924 \, \text{m}\)` above its initial position.

How did we get the value?

Let's do the calculations:

Given values:

- L = 20.0 cm

- W = 15.0 cm

- D = 3.0 cm

- k = 800 N/m

- m = 0.150 kg

- s = 6.00 cm

- P = 24.6 kPa

-
`\( g = 9.8 \, \text{m/s}^2 \)`

1. Calculate x :

A = L x W

`\[ x = \sqrt{(P \cdot A)/(k)} \]`

`\[ A = (0.20 \, \text{m}) * (0.15 \, \text{m}) \]`

`\[ A = 0.03 \, \text{m}^2 \]`

`\[ x = \sqrt{\frac{(24.6 * 10^3 \, \text{Pa}) * (0.03 \, \text{m}^2)}{800 \, \text{N/m}}} \]`

Calculating x:

`\[ x \approx 0.108 \, \text{m} \]`

2. Calculate v:

`\[ v = \sqrt{(k)/(m) x} \]`

`\[ v = \sqrt{\frac{800 \, \text{N/m}}{0.150 \, \text{kg}} * 0.108 \, \text{m}} \]`

Calculating v:

`\[ v \approx 4.16 \, \text{m/s} \]`

3. Calculate t:

`\[ t = (v)/(g) \]`

`\[ t = \frac{4.16 \, \text{m/s}}{9.8 \, \text{m/s}^2} \]`

Calculating t:

`\[ t \approx 0.425 \, \text{s} \]`

4. Calculate h:

`\[ h = (1)/(2) g t^2 \]`

`\[ h = (1)/(2) * 9.8 \, \text{m/s}^2 * (0.425 \, \text{s})^2 \]`

Calculating h:

`\[ h \approx 0.924 \, \text{m} \]`

Therefore, the maximum height gained by the cube is approximately
`\(0.924 \, \text{m}\)` above its initial position.