Question

practice it
use the worked example above to help you solve this problem. a 49.5 kg circus acrobat drops from a height of 2.25 meters straight down onto a springboard with a force constant of 8.10×10³ n/m, as shown in the figure. by what maximum distance does she compress the spring?
0.51
your response differs from the correct answer by more than 10%. double check your calculations. m
exercise
hints: getting started | im stuck!
an 6.52 kg block drops straight down from a height of 1.14 m, striking a platform spring having a force constant of 1.16×10³ n/m. find the maximum compression of the spring.
0.36
do not neglect the change in gravitational potential energy after the block hits the spring. m

Explanation:

Step1: Set up energy - conservation equation

The initial gravitational potential energy of the object is converted into elastic potential energy of the spring and gravitational potential energy during the compression of the spring. Let the mass of the object be $m$, the initial height above the un - compressed spring be $h$, the spring constant be $k$, and the compression of the spring be $x$. The initial gravitational potential energy is $U_{g1}=mg(h + x)$, and the elastic potential energy of the spring is $U_{s}=\frac{1}{2}kx^{2}$. So, $mg(h + x)=\frac{1}{2}kx^{2}$.

Step2: Rearrange the equation into a quadratic form

For the first problem with $m = 49.5$ kg, $h=2.25$ m, and $k = 8.10\times10^{3}$ N/m, we have $49.5\times9.8\times(2.25 + x)=\frac{1}{2}\times8.10\times10^{3}x^{2}$.
Expanding gives $49.5\times9.8\times2.25+49.5\times9.8x = 4050x^{2}$.
$1089.525+485.1x = 4050x^{2}$.
$4050x^{2}-485.1x - 1089.525 = 0$.
Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 4050$, $b=-485.1$, and $c=-1089.525$.
$x=\frac{485.1\pm\sqrt{(-485.1)^{2}-4\times4050\times(-1089.525)}}{2\times4050}$.
$x=\frac{485.1\pm\sqrt{235322.01+17573220}}{8100}$.
$x=\frac{485.1\pm\sqrt{17808542.01}}{8100}$.
$x=\frac{485.1\pm4219.9}{8100}$. We take the positive root $x=\frac{485.1 + 4219.9}{8100}=\frac{4705}{8100}\approx0.58$ m.
For the second problem with $m = 6.52$ kg, $h = 1.14$ m, and $k = 1.16\times10^{3}$ N/m.
$mg(h + x)=\frac{1}{2}kx^{2}$.
$6.52\times9.8\times(1.14 + x)=\frac{1}{2}\times1160x^{2}$.
$71.47\times(1.14 + x)=580x^{2}$.
$81.4758+71.47x = 580x^{2}$.
$580x^{2}-71.47x - 81.4758 = 0$.
Using the quadratic formula with $a = 580$, $b=-71.47$, and $c=-81.4758$.
$x=\frac{71.47\pm\sqrt{(-71.47)^{2}-4\times580\times(-81.4758)}}{2\times580}$.
$x=\frac{71.47\pm\sqrt{5108.96+189870.34}}{1160}$.
$x=\frac{71.47\pm\sqrt{194979.3}}{1160}$.
$x=\frac{71.47\pm441.56}{1160}$. We take the positive root $x=\frac{71.47 + 441.56}{1160}=\frac{513.03}{1160}\approx0.44$ m.

Answer:

For the first problem: $0.58$ m
For the second problem: $0.44$ m