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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 stikes to the spring, which is now compressed 1.5 x 10⁻² m bringing the block to a momentary halt. assume air resistance is negligible.
what is the elastic potential energy of the compressed spring?
7.00 j
3.94 x 10⁻² j
4.33 x 10⁻³ j
1.05 j

Explanation:

Step1: Recall elastic potential energy formula

Elastic potential energy \( U = \frac{1}{2}kx^2 \), where \( k=350\,\text{N/m} \), \( x=1.5\times10^{-2}\,\text{m} \).

Step2: Substitute values into formula

Calculate \( x^2 = (1.5\times10^{-2})^2 = 2.25\times10^{-4}\,\text{m}^2 \). Then \( U = \frac{1}{2}\times350\times2.25\times10^{-4} \).

Step3: Compute the result

\( \frac{1}{2}\times350 = 175 \); \( 175\times2.25\times10^{-4} = 393.75\times10^{-4} = 3.9375\times10^{-2}\,\text{J} \approx 3.94\times10^{-2}\,\text{J} \).

Answer:

B. \( 3.94\times10^{-2}\,\text{J} \)