Question

a 1.2kg physics cart is initially travelling 2.4m/s when it collides with an ideal spring (with a spring constant of 790n/m and natural length shown). the final time is defined as when the spring at maximum compression and the cart is about to rebound backwards. neglect loss. list the energy types at the initial and final time and whether work and loss (due to non - conservative forces) occur as well as the corresponding amounts of energy. include the earth as part of the system. initial: ke = 3.46 j work? no work = 0 j loss? no loss = 0 j final: epe = 3.46 j how much did the springs length change during the compression? m what is the length of the spring at maximum compression? m what is the spring force at this time? n

Explanation:

Step1: Analyze Energy Conservation

The cart's initial kinetic energy (KE) converts to elastic potential energy (EPE) at maximum compression (since velocity is 0 there). Initial KE: $KE = \frac{1}{2}mv^2$. Given $m = 1.2\,\text{kg}$, $v = 2.4\,\text{m/s}$, so $KE = 0.5 \times 1.2 \times (2.4)^2 = 3.456 \approx 3.46\,\text{J}$, matching the given value.

Step2: Find Spring Compression (Δx)

At maximum compression, $EPE = \frac{1}{2}k\Delta x^2$. We know $EPE = 3.46\,\text{J}$, $k = 790\,\text{N/m}$. Solve for $\Delta x$:
$\Delta x = \sqrt{\frac{2 \times EPE}{k}} = \sqrt{\frac{2 \times 3.46}{790}} \approx \sqrt{\frac{6.92}{790}} \approx \sqrt{0.00876} \approx 0.0936\,\text{m}$? Wait, no—wait, the natural length is 0.375m? Wait, maybe the "length at max compression" is natural length minus compression? Wait, no, the question "How much did the spring’s length change" is compression (Δx). Wait, let's recheck: $EPE = \frac{1}{2}k\Delta x^2 \implies \Delta x = \sqrt{\frac{2EPE}{k}}$. Plugging $EPE = 3.46$, $k = 790$:
$\Delta x = \sqrt{\frac{2 \times 3.46}{790}} = \sqrt{\frac{6.92}{790}} \approx \sqrt{0.00876} \approx 0.0936\,\text{m}$? But maybe the natural length is 0.375m (from the diagram). Wait, the next question: "length at max compression" is natural length - Δx? Wait, no, the change in length is Δx (compression), so the length change is Δx. Wait, let's recalculate:
$EPE = 3.46 = 0.5 \times 790 \times \Delta x^2 \implies \Delta x^2 = \frac{3.46 \times 2}{790} = \frac{6.92}{790} \approx 0.00876 \implies \Delta x \approx \sqrt{0.00876} \approx 0.0936\,\text{m} \approx 0.094\,\text{m}$? Wait, maybe the diagram shows natural length 0.375m, so max compression length is 0.375 - Δx? No, the question "How much did the spring’s length change" is the compression (Δx), so that's the change.

Step3: Length at Max Compression

Natural length (L₀) is 0.375m (from diagram). Length at max compression: $L = L₀ - \Delta x$. We found Δx ≈ 0.0936m, so $L = 0.375 - 0.0936 ≈ 0.281\,\text{m}$? Wait, no—wait, maybe I mixed up. Wait, the spring's natural length is 0.375m (the right diagram shows 0.375m as reference). When compressed, the length decreases by Δx. So length change is Δx (compression), so length at max compression is 0.375 - Δx.

Step4: Spring Force at Max Compression

Hooke's Law: $F = k\Delta x$. $k = 790\,\text{N/m}$, Δx ≈ 0.0936m, so $F = 790 \times 0.0936 ≈ 73.9\,\text{N}$? Wait, but let's use EPE to find Δx correctly. $EPE = 3.46 = 0.5 \times 790 \times \Delta x^2 \implies \Delta x = \sqrt{\frac{2 \times 3.46}{790}} = \sqrt{\frac{6.92}{790}} ≈ \sqrt{0.00876} ≈ 0.0936\,\text{m}$. Then $F = k\Delta x = 790 \times 0.0936 ≈ 73.9\,\text{N} ≈ 74\,\text{N}$? Wait, but maybe the initial calculation for Δx is wrong. Wait, let's check KE again: $0.5 \times 1.2 \times (2.4)^2 = 0.5 \times 1.2 \times 5.76 = 3.456\,\text{J}$, correct. Then $EPE = 3.456 = 0.5 \times 790 \times \Delta x^2 \implies \Delta x^2 = (3.456 \times 2)/790 = 6.912/790 ≈ 0.00875\implies \Delta x ≈ \sqrt{0.00875} ≈ 0.0935\,\text{m} ≈ 0.094\,\text{m}$. So length change (compression) is ~0.094m. Length at max compression: 0.375 - 0.094 ≈ 0.281m. Spring force: $F = 790 \times 0.0935 ≈ 73.9\,\text{N} ≈ 74\,\text{N}$.

Answer:

• How much did the spring’s length change: $\boldsymbol{\approx 0.094\,\text{m}}$
• Length at max compression: $\boldsymbol{\approx 0.281\,\text{m}}$ (or 0.375 - 0.094)
• Spring force: $\boldsymbol{\approx 74\,\text{N}}$ (using $F = k\Delta x = 790 \times 0.0935$)

(Note: Depending on rounding, values may vary slightly. The key steps use energy conservation, Hooke's Law, and length change as compression.)