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description\tshow all you work, including units, on separate paper. follow the \problem solving method\.
\tthis final exam is cumulative and covers material from the entire course.
instructions\tfrom the list of choices, select the one best answer.
multiple attempts\tnot allowed. this test can only be taken once.
force completion\tthis test can be saved and resumed later.
\tyour answers are saved automatically
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question 57
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a 0.10 kg ball rests on a compressed spring. the spring is compressed 0.25 m from its unstrained length. when the spring is released, the ball rises to a maximum height of 8.0 m above its position on the compressed spring. ignoring air resistance, determine the spring constant.
302 n/m
251 n/m
85 n/m
170 n/m

Explanation:

Step1: Identify Energy Conservation

Use conservation of mechanical energy: elastic potential energy in spring converts to gravitational potential energy of ball. Formulas: $\frac{1}{2}kx^2 = mgh$.

Step2: List Known Values

Mass $m = 0.10\ kg$, compression $x = 0.25\ m$, height $h = 8.0\ m$, $g = 9.8\ m/s^2$.

Step3: Solve for k

Rearrange formula: $k=\frac{2mgh}{x^2}$. Substitute values: $k=\frac{2\times0.10\times9.8\times8.0}{(0.25)^2}$. Calculate numerator: $2\times0.10\times9.8\times8.0 = 15.68$. Denominator: $(0.25)^2 = 0.0625$. Then $k=\frac{15.68}{0.0625}=250.88\approx251\ N/m$.

Answer:

251 N/m (corresponding to the option "251 N/m")