Given:

• Stretch distance, $x = 3.00 \, \text{cm} = 0.0300 \, \text{m}$
• Work done to stretch, $W = 12.0 \, \text{J}$

(a) Force constant of the spring

The work done to stretch a spring is related to the spring constant $k$ and the displacement $x$ by the formula:

$W = \frac{1}{2} k x^2$

We can rearrange this equation to solve for $k$:

$k = \frac{2W}{x^2}$

Substitute the values:

$k = \frac{2 \times 12.0 \, \text{J}}{(0.0300 \, \text{m})^2}$

$k = \frac{24.0 \, \text{J}}{0.0009 \, \text{m}^2}$

$k = 26,666.67 \, \text{N/m}$

So, the force constant $k$ is approximately $26,667 \, \text{N/m}$.

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(b) Force needed to stretch the spring 3.00 cm

The force $F$ needed to stretch or compress a spring is given by Hooke's law:

$F = kx$

Substituting the values of $k$ and $x$:

$F = 26,667 \, \text{N/m} \times 0.0300 \, \text{m}$

$F = 800.0 \, \text{N}$

So, the force needed to stretch the spring 3.00 cm is $800.0 \, \text{N}$.

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(c) Work and force needed to compress the spring 4.00 cm

1. Work done to compress the spring 4.00 cm:

We use the same work formula, but with $x = 4.00 \, \text{cm} = 0.0400 \, \text{m}$:

$W = \frac{1}{2} k x^2$

Substitute the values:

$W = \frac{1}{2} \times 26,667 \, \text{N/m} \times (0.0400 \, \text{m})^2$

$W = 0.5 \times 26,667 \times 0.0016 \, \text{m}^2$

$W = 21.33 \, \text{J}$

So, the work required to compress the spring 4.00 cm is $21.33 \, \text{J}$.

2. Force needed to compress the spring 4.00 cm:

We use Hooke’s law again:

$F = kx$

Substitute the values:

$F = 26,667 \, \text{N/m} \times 0.0400 \, \text{m}$

$F = 1,066.68 \, \text{N}$

So, the force needed to compress the spring 4.00 cm is approximately $1,067 \, \text{N}$.

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Summary:

• (a) Force constant $k = 26,667 \, \text{N/m}$
• (b) Force to stretch 3.00 cm: $800.0 \, \text{N}$
• (c) Work to compress 4.00 cm: $21.33 \, \text{J}$, Force to compress 4.00 cm: $1,067 \, \text{N}$

Would you like more details or have any questions?

Related Questions:

1. How does the work done change if the spring is stretched by 6.00 cm instead of 3.00 cm?
2. What would the force constant be if the work done to stretch the spring by 3.00 cm were doubled?
3. How does the energy stored in a spring change when it is compressed instead of stretched?
4. If a different spring has a force constant of 50,000 N/m, what force would be required to stretch it by 2.00 cm?
5. How does the Hooke’s law equation change when considering a spring system in parallel?

Tip: For springs, the work done is proportional to the square of the displacement, meaning small changes in stretch or compression result in significant changes in energy stored.