Question

remarks the other solution, d = -0.437 m, can be rejected because d was chosen to be a positive number at the outset. a change in the acrobats center of mass, say, by crouching as she makes contact with the springboard, also affects the springs compression, but that effect was neglected. shock absorbers often involve springs, and this example illustrates how they work. the spring action of a shock absorber turns a dangerous jolt into a smooth deceleration, as excess kinetic energy is converted to spring potential energy. question is it possible for the acrobat to rebound to a height greater than her initial height? explain. (select all that apply.) no. the kinetic energy that the acrobat gains on the way down is converted entirely back into potential energy when she reaches the initial height. yes. elastic energy is always present in the spring and can give the acrobat greater height than initially. yes. the acrobat can bend her knees while falling and then straighten them as if jumping when bouncing upward again. yes. the acrobat can provide mechanical energy by pushing herself up while in contact with the springboard. no. there is no external source of energy to provide the potential energy at a greater height. consider whether the acrobat standing on the ground, without any spring at all, could jump to a greater height than she had initially when simply standing in place. then apply the same idea to what the acrobat could try on a trampoline. 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 x your response differs from the correct answer by more than 10%. double check your calculations. m

Explanation:

Step1: Apply conservation of mechanical energy

The initial gravitational - potential energy of the acrobat is converted into elastic - potential energy of the spring and gravitational - potential energy at the compressed position. The initial gravitational potential energy of the acrobat is $U_{g1}=mgh$, where $m = 49.5$ kg, $g = 9.8$ m/s², and $h = 2.25$ m. At the maximum compression $x$ of the spring, the total mechanical energy is $U_{g2}+U_{s}$, where $U_{g2}=mg( - x)$ (taking downwards as negative) and $U_{s}=\frac{1}{2}kx^{2}$, with $k = 8.10\times10^{3}$ N/m.
So, $mgh=\frac{1}{2}kx^{2}-mgx$.

Step2: Rearrange the equation to a quadratic form

We get $\frac{1}{2}kx^{2}-mgx - mgh=0$. Substituting the values: $\frac{1}{2}\times8.10\times10^{3}x^{2}-49.5\times9.8x-49.5\times9.8\times2.25 = 0$.
$4050x^{2}-485.1x - 1081.425 = 0$.

Step3: Use the quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 4050$, $b=-485.1$, and $c=-1081.425$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-485.1)^{2}-4\times4050\times(-1081.425)$.
$\Delta = 235322.01+17479470=17714792.01$.
Then, $x=\frac{485.1\pm\sqrt{17714792.01}}{8100}$.
$x=\frac{485.1\pm4208.9}{8100}$.
We take the positive root $x=\frac{485.1 + 4208.9}{8100}=\frac{4694}{8100}\approx0.58$ m.

Answer:

$0.58$ m