Question

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. 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? 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. m

Explanation:

Step1: Consider conservation of mechanical energy

The total mechanical energy of the system (acrobat - spring - Earth) is conserved in the absence of non - conservative forces (like air resistance). When the acrobat falls, her initial gravitational potential energy is converted into kinetic energy and then into elastic potential energy of the spring. When she rebounds, the elastic potential energy of the spring is converted back into kinetic energy and then gravitational potential energy.

If there is no external work done on the system, the maximum height she can reach is her initial height because the kinetic energy she gains on the way down is converted entirely back into potential energy when she reaches the initial height. So, "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." is correct.

Also, there is no external source of energy to provide the potential energy at a greater height in a conservative - force only situation. So, "No. There is no external source of energy to provide the potential energy at a greater height." is correct.

The statements about elastic energy always being present to give greater height, the acrobat bending and straightening knees, and the acrobat providing mechanical energy by pushing herself up are incorrect because they imply an increase in total mechanical energy of the system without an external source.

Step2: For the practice it and exercise parts

We use the conservation of mechanical energy. The initial gravitational potential energy of the object is converted into elastic potential energy of the spring at maximum compression. Let the mass of the object be $m$, the initial height be $h$, the spring - force constant be $k$, and the maximum compression of the spring be $x$.

The initial gravitational potential energy is $U_{g}=mgh$ and the elastic potential energy of the spring is $U_{s}=\frac{1}{2}kx^{2}$. At maximum compression, $mgh=\frac{1}{2}kx^{2}$.

For the practice it problem:
$m = 49.5\ kg$, $h = 2.25\ m$, $k=8.10\times 10^{3}\ N/m$
$mgh=\frac{1}{2}kx^{2}$
$x=\sqrt{\frac{2mgh}{k}}$
$x=\sqrt{\frac{2\times49.5\times9.8\times2.25}{8.10\times 10^{3}}}$
$x=\sqrt{\frac{2\times49.5\times9.8\times2.25}{8100}}$
$x=\sqrt{\frac{2\times49.5\times9.8\times2.25}{81\times100}}$
$x=\sqrt{\frac{2\times49.5\times9.8\times2.25}{81}\times\frac{1}{100}}$
$x=\sqrt{\frac{2\times49.5\times9.8\times2.25}{81}}\times\frac{1}{10}$
$x=\sqrt{\frac{2\times49.5\times9.8\times2.25}{81}}\times0.1$
$x=\sqrt{\frac{2\times49.5\times9.8\times2.25}{81}}\approx0.51\ m$

For the exercise problem:
$m = 6.52\ kg$, $h = 1.14\ m$, $k = 1.16\times 10^{3}\ N/m$
$x=\sqrt{\frac{2mgh}{k}}$
$x=\sqrt{\frac{2\times6.52\times9.8\times1.14}{1.16\times 10^{3}}}$
$x=\sqrt{\frac{2\times6.52\times9.8\times1.14}{1160}}$
$x\approx0.36\ m$

Answer:

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.
No. There is no external source of energy to provide the potential energy at a greater height.
Practice it: $0.51\ m$
Exercise: $0.36\ m$