Question
A spring extends by 2cm when a mass of 40 grams is suspended on it.What is the weight required to extend it by 2.5cm
Answers
GPT-4o mini
To find the weight required to extend the spring by 2.5 cm, we can first determine the spring constant \( k \) using Hooke’s law, which states that the force applied to a spring is proportional to its extension. Hooke's law is given by:
\[
F = k \cdot x
\]
where:
- \( F \) is the force (weight) applied,
- \( k \) is the spring constant,
- \( x \) is the extension (displacement from the equilibrium position).
First, we determine the force applied by the 40 grams mass.
1. **Convert mass to weight** (force):
\[
\text{Mass} = 40 \text{ grams} = 0.04 \text{ kg}
\]
\[
F = m \cdot g = 0.04 \text{ kg} \cdot 9.81 \text{ m/s}^2 \approx 0.3924 \text{ N}
\]
2. **Calculate spring constant \( k \)** with the original extension of 2 cm (0.02 m):
\[
x_1 = 0.02 \text{ m}
\]
\[
k = \frac{F}{x_1} = \frac{0.3924 \text{ N}}{0.02 \text{ m}} = 19.62 \text{ N/m}
\]
3. **Determine the force required to extend it by 2.5 cm (0.025 m)** using the spring constant \( k \):
\[
x_2 = 0.025 \text{ m}
\]
\[
F_2 = k \cdot x_2 = 19.62 \text{ N/m} \cdot 0.025 \text{ m} = 0.4905 \text{ N}
\]
4. **Convert this force back to mass**:
\[
\text{Weight} = F_2 = 0.4905 \text{ N}
\]
\[
\text{Mass} = \frac{F_2}{g} = \frac{0.4905 \text{ N}}{9.81 \text{ m/s}^2} \approx 0.05 \text{ kg} \approx 50 \text{ grams}
\]
Therefore, the weight required to extend the spring by 2.5 cm is approximately **50 grams**.
\[
F = k \cdot x
\]
where:
- \( F \) is the force (weight) applied,
- \( k \) is the spring constant,
- \( x \) is the extension (displacement from the equilibrium position).
First, we determine the force applied by the 40 grams mass.
1. **Convert mass to weight** (force):
\[
\text{Mass} = 40 \text{ grams} = 0.04 \text{ kg}
\]
\[
F = m \cdot g = 0.04 \text{ kg} \cdot 9.81 \text{ m/s}^2 \approx 0.3924 \text{ N}
\]
2. **Calculate spring constant \( k \)** with the original extension of 2 cm (0.02 m):
\[
x_1 = 0.02 \text{ m}
\]
\[
k = \frac{F}{x_1} = \frac{0.3924 \text{ N}}{0.02 \text{ m}} = 19.62 \text{ N/m}
\]
3. **Determine the force required to extend it by 2.5 cm (0.025 m)** using the spring constant \( k \):
\[
x_2 = 0.025 \text{ m}
\]
\[
F_2 = k \cdot x_2 = 19.62 \text{ N/m} \cdot 0.025 \text{ m} = 0.4905 \text{ N}
\]
4. **Convert this force back to mass**:
\[
\text{Weight} = F_2 = 0.4905 \text{ N}
\]
\[
\text{Mass} = \frac{F_2}{g} = \frac{0.4905 \text{ N}}{9.81 \text{ m/s}^2} \approx 0.05 \text{ kg} \approx 50 \text{ grams}
\]
Therefore, the weight required to extend the spring by 2.5 cm is approximately **50 grams**.
oobleck
or, since F = kx, just solve
40/2 = x/2.5
x = 50
40/2 = x/2.5
x = 50
GPT-4o mini
Yes, that's a simpler way to arrive at the answer using a proportion since we know that the force exerted by the spring is directly proportional to the extension according to Hooke's Law (\( F = kx \)).
From the information provided:
- When the weight of 40 grams (0.04 kg) produces an extension of 2 cm (0.02 m), we can establish the ratio:
\[
\frac{F_1}{x_1} = \frac{F_2}{x_2}
\]
Substituting the known values:
\[
\frac{40 \text{ grams}}{2 \text{ cm}} = \frac{x \text{ grams}}{2.5 \text{ cm}}
\]
Cross-multiplying, we get:
\[
40 \text{ grams} \times 2.5 \text{ cm} = x \text{ grams} \times 2 \text{ cm}
\]
\[
100 = 2x \implies x = 50 \text{ grams}
\]
Thus, using this proportional reasoning, you correctly arrive at the answer that **50 grams** is required to extend the spring by 2.5 cm. Thank you for pointing out the simplification!
From the information provided:
- When the weight of 40 grams (0.04 kg) produces an extension of 2 cm (0.02 m), we can establish the ratio:
\[
\frac{F_1}{x_1} = \frac{F_2}{x_2}
\]
Substituting the known values:
\[
\frac{40 \text{ grams}}{2 \text{ cm}} = \frac{x \text{ grams}}{2.5 \text{ cm}}
\]
Cross-multiplying, we get:
\[
40 \text{ grams} \times 2.5 \text{ cm} = x \text{ grams} \times 2 \text{ cm}
\]
\[
100 = 2x \implies x = 50 \text{ grams}
\]
Thus, using this proportional reasoning, you correctly arrive at the answer that **50 grams** is required to extend the spring by 2.5 cm. Thank you for pointing out the simplification!