To calculate the spring constant $k$ for the spring, we use Hooke's law, which is expressed as:

$F = -k x$

Where:

• $F$ is the force exerted by the spring (in Newtons),
• $k$ is the spring constant (in N/m),
• $x$ is the displacement of the spring from its equilibrium position (in meters).

Since the spring is stretched by a mass $m$ hanging from it, the force $F$ applied by the mass is due to gravity. This force is given by:

$F = mg$

Where:

• $m = 0.5 \, \text{kg}$ is the mass,
• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

Now, we know the force $F$ and the displacement $x$, so we can solve for the spring constant $k$.

1. Calculate the force: $F = mg = (0.5 \, \text{kg})(9.8 \, \text{m/s}^2) = 4.9 \, \text{N}$

2. Use Hooke's law $F = kx$ to solve for $k$: $k = \frac{F}{x} = \frac{4.9 \, \text{N}}{0.07 \, \text{m}} = 70 \, \text{N/m}$

Thus, the spring constant $k$ is 70 N/m.

Would you like a more detailed explanation or have any other questions?

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Here are 5 related questions:

1. How would the spring constant change if the spring stretched 0.1 meters instead?
2. What happens to the spring constant if a mass of 1 kg is used instead of 0.5 kg?
3. Can Hooke's law be applied if the spring is compressed instead of stretched? Why?
4. How would the force change if the acceleration due to gravity were on another planet, say Mars (3.7 m/s²)?
5. If a different spring had a constant of 140 N/m, how much would it stretch with a 0.5 kg mass?

Tip: The spring constant $k$ is a property of the spring itself—it doesn't change with the mass hanging from it or the gravitational field but rather reflects how stiff the spring is.