Question

1. a certain spring that obeys hooke’s law is stretched by an external agent. the work done in stretching the spring by 10 cm is 4 j. how much additional work is required to stretch the spring an additional 10 cm? (a) 2 j (b) 4 j (c) 8 j (d) 12 j (e) 16 j

Explanation:

Step1: Recall Work Formula for Spring

The work done on a spring obeying Hooke's law is $W = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement.

Step2: Find $k$ from First Stretch

Given $W_1 = 4\,\text{J}$ and $x_1 = 10\,\text{cm} = 0.1\,\text{m}$. Substitute into $W_1 = \frac{1}{2}kx_1^2$:
$4 = \frac{1}{2}k(0.1)^2$
Solve for $k$:
$k = \frac{4 \times 2}{(0.1)^2} = 800\,\text{N/m}$.

Step3: Calculate Work for Total Stretch

Total stretch $x_2 = 10 + 10 = 20\,\text{cm} = 0.2\,\text{m}$.
Work $W_2 = \frac{1}{2}kx_2^2 = \frac{1}{2} \times 800 \times (0.2)^2 = 16\,\text{J}$.

Step4: Find Additional Work

Additional work $W_{\text{additional}} = W_2 - W_1 = 16 - 4 = 12\,\text{J}$? Wait, no—wait, wait, mistake here. Wait, first stretch is 10 cm (0.1 m) with work 4 J. Then stretching an additional 10 cm means total stretch is 20 cm (0.2 m). Wait, recalculate $W_2$:
Wait, $\frac{1}{2}k(0.2)^2 = \frac{1}{2}k(0.04)$, and $\frac{1}{2}k(0.01) = 4\,\text{J}$. So $\frac{1}{2}k(0.04) = 4 \times 4 = 16\,\text{J}$ (since 0.04 is 4 times 0.01). Then additional work is $16 - 4 = 12$? Wait, no, wait the options have 12? Wait no, wait the options are (a)2, (b)4, (c)8, (d)12, (e)16. Wait, no—wait, maybe I messed up units. Wait, $x$ in cm: let's keep $x$ in cm for ratio.
Since $W \propto x^2$, for $x_1 = 10\,\text{cm}$, $W_1 = 4\,\text{J}$. For $x_2 = 20\,\text{cm}$, $W_2 = 4 \times (\frac{20}{10})^2 = 4 \times 4 = 16\,\text{J}$. Then additional work is $16 - 4 = 12\,\text{J}$? Wait, but wait the problem says "stretch the spring an additional 10 cm"—so from 10 cm to 20 cm. So yes, $W_2 - W_1 = 16 - 4 = 12\,\text{J}$. Wait, but let me check again. Wait, no—wait, maybe the initial calculation was wrong. Wait, $x_1 = 0.1\,\text{m}$, $W_1 = 4 = 0.5 \times k \times 0.01$ → $k = 800$. Then $x_2 = 0.2\,\text{m}$, $W_2 = 0.5 \times 800 \times 0.04 = 16$. Then 16 - 4 = 12. So additional work is 12 J? But wait, the options have (d)12 J. Wait, but let me check the ratio again. Since $W \propto x^2$, so when $x$ doubles (from 10 to 20 cm), $W$ becomes 4 times (since $2^2 = 4$). So initial $W$ at 10 cm: 4 J. At 20 cm: 16 J. Additional work: 16 - 4 = 12 J. Yes, that's correct.

Answer:

(d) 12 J