To determine how much a 25-gram mass would stretch the same spring, we can use Hooke's Law, which states that the force exerted by the spring is proportional to the displacement (stretch) of the spring.
The relationship can be represented as: \[ F = kx \] where \( F \) is the force exerted (in this case, due to the weight of the mass), \( k \) is the spring constant, and \( x \) is the displacement (stretch) of the spring.
The weight of an object can be calculated using: \[ F = mg \] where \( m \) is the mass in kilograms and \( g \) is the acceleration due to gravity (approximately \( 9.8 , \text{m/s}^2 \)).
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For the 100-gram mass (0.1 kg): \[ F_{100} = 0.1 , \text{kg} \times 9.8 , \text{m/s}^2 = 0.98 , \text{N} \] This causes the spring to stretch 8 cm (0.08 m).
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Now, for the 25-gram mass (0.025 kg): \[ F_{25} = 0.025 , \text{kg} \times 9.8 , \text{m/s}^2 = 0.245 , \text{N} \]
Now we observe that the force exerted by the 25-gram mass is proportional to the displacement we are interested in. Since the 100-gram mass stretched the spring 8 cm: \[ \frac{F_{25}}{F_{100}} = \frac{x_{25}}{x_{100}} \]
Plugging in the values: \[ \frac{0.245 , \text{N}}{0.98 , \text{N}} = \frac{x_{25}}{0.08 , \text{m}} \]
Solving for \( x_{25} \): \[ x_{25} = 0.08 , \text{m} \times \frac{0.245}{0.98} \approx 0.02 , \text{m} = 2 , \text{cm} \]
Thus, a 25-gram mass would stretch the spring 2 cm.
The correct response is: Stretch the spring 2 cm.