Question

a 0.200 kg block is held in place by a force \\( \vec{f} \\) that results in a 0.100 m compression of a spring beneath the block. the spring constant is \\( 1.0 \times 10^2 \\) n/m. assume that the mass of the spring is negligible compared to that of the block. what is the elastic potential energy of the compressed spring?
\\( \bigcirc \\) 2.00 j
\\( \bigcirc \\) 0.200 j
\\( \bigcirc \\) 0.500 j
\\( \bigcirc \\) 1.00 j

Explanation:

Step1: Recall elastic potential energy formula

Elastic potential energy $U = \frac{1}{2}kx^2$, where $k$ is spring constant, $x$ is compression.

Step2: Substitute given values

$k = 1.0 \times 10^2$ N/m, $x = 0.100$ m.
$U = \frac{1}{2} \times 100 \times (0.100)^2 = 0.500$ J.

Answer:

0.500 J