Question

problem reference 10.3
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?
to what maximum height would the block rise if the force \\(\vec{f}\\) were removed?

\\(\bigcirc\\) 5.00 m
\\(\bigcirc\\) 0.520 m
\\(\bigcirc\\) 0.255 m
\\(\bigcirc\\) 2.50 m

Explanation:

Step1: Calculate elastic potential energy

Elastic potential energy formula: $U = \frac{1}{2}kx^2$.
$k=1.0×10^2$ N/m, $x=0.100$ m.
$U = \frac{1}{2}×100×(0.1)^2 = 0.5$ J.

Step2: Relate potential energy to height

Energy conservation: $U = mgh$.
Solve for $h$: $h = \frac{U}{mg}$.
$m=0.200$ kg, $g=9.8$ m/s².
$h = \frac{0.5}{0.2×9.8} ≈ 0.255$ m.

Answer:

Elastic potential energy: 0.5 J
Maximum height: 0.255 m
(For the multiple-choice part: 0.255 m)