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question 17
problem reference 10.4
a vertical spring with a spring constant of 350 n/m is mounted on the floor. from directly above the spring, which is unstrained, a 0.200 kg block is dropped from rest. it collides and sticks to the spring, which is now compressed 1.5 × 10⁻² m bringing the block to a momentary halt. assume air resistance is negligible.
from what height above the compressed spring was the block dropped?
0.870 m
6.67 × 10⁻³ m
5.90 × 10⁻² m
2.01 × 10⁻² m

Explanation:

Step1: Define variables

Let \( h \) = height above compressed spring, \( m=0.200\,\text{kg} \), \( k=350\,\text{N/m} \), \( x=1.5×10^{-2}\,\text{m} \), \( g=9.8\,\text{m/s}^2 \).

Step2: Apply energy conservation

Gravitational potential energy = Spring potential energy:
\( mg(h + x) = \frac{1}{2}kx^2 \)

Step3: Solve for \( h \)

Rearrange: \( h = \frac{kx^2}{2mg} - x \)
Substitute values:
\( \frac{350×(1.5×10^{-2})^2}{2×0.200×9.8} - 1.5×10^{-2} \)
Calculate numerator: \( 350×2.25×10^{-4}=0.07875 \)
Denominator: \( 2×0.2×9.8=3.92 \)
First term: \( 0.07875/3.92≈0.02009 \)
Subtract \( x \): \( 0.02009 - 0.015≈0.00509\,\text{m}≈5.09×10^{-3}\,\text{m} \)? Wait correction: Wait \( (1.5×10^{-2})^2=2.25×10^{-4} \), 350×2.25e-4=0.07875; 20.29.8=3.92; 0.07875/3.92≈0.02009m=2.009×10^{-2}m; minus 0.015m=0.00509m≈5.1×10^{-3}m? Wait no wait original question says "height above the compressed spring"—wait wait: If the spring is compressed by \( x \), then the distance fallen is \( h + x \) where \( h \) is above compressed spring. So \( mg(h + x) = 1/2 kx² \) → \( h = (kx²)/(2mg) - x \). Let's recalculate precisely:
\( kx²=350(0.015)^2=3500.000225=0.07875 \)
\( 2mg=2*0.2*9.8=3.92 \)
\( kx²/(2mg)=0.07875/3.92≈0.020089 \)
\( h=0.020089 - 0.015=0.005089m≈5.09×10^{-3}m≈5.1×10^{-3}m \)? But the options have 6.67e-3, 5.90e-2, 2.01e-2. Wait wait—wait did I misread the compression? The problem says "compressed 1.5×10⁻² m"=0.015m. Wait wait another way: Maybe the question means height above unstrained spring? No, it says "above the compressed spring". Wait wait let's check the option 2.01×10⁻² m: that's the value of \( kx²/(2mg)≈0.0201m=2.01×10^{-2}m \). Oh! Wait maybe I misinterpreted the height: If "height above compressed spring" is the distance from release point to compressed position, then \( h \) is that distance, so \( mgh=1/2kx² \)? Wait no—because when you drop from above compressed spring, the block falls distance \( h \) (which is the height above compressed spring) to reach the compressed position, so the fall distance is \( h \), so \( mgh=1/2kx² \). Let's calculate that: \( h=(kx²)/(2mg)=0.07875/3.92≈0.02009m≈2.01×10^{-2}m \). Oh! That's option D: 2.01×10⁻² m. I made a mistake earlier by adding \( x \)—no, if the height is above the compressed spring, then the total fall distance is exactly that height (since compressed spring is the final position), so no need to add \( x \). That's the error.

Step4: Correct calculation

\( h = \frac{kx²}{2mg} = \frac{350(1.5×10^{-2})²}{20.200*9.8} ≈ 0.0201m = 2.01×10^{-2}m \)

Answer:

D. 2.01 x 10⁻² m