**LEARN MORE****REMARKS** The other solution, \( d = -0.437 \, \text{m} \), can be rejected because \( d \) was chosen to be a positive number at the outset. A change in the acrobat's center of mass, say, by crouching as she makes contact with the springboard, also affects the spring's 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. The acrobat can provide mechanical energy by pushing herself up while in contact with the springboard.- □ 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.- □ 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 \( 49.7 \, \text{kg} \) circus acrobat drops from a height of \( 1.57 \, \text{meters} \) straight down onto a springboard with a force constant of \( 6.90 \times 10^3 \, \text{N/m} \), as shown in the figure. By what maximum distance does she compress the spring?\[\underline{\hphantom{mm}} \, \text{m}\]---**EXERCISE****HINTS:** [GETTING STARTED](#) | [I'M STUCK!](#)An \( 6.37 \, \text{kg} \) block drops straight down from a height of \( 0.77 \, \text{m} \), striking a platform spring having a force constant of \( 9.50 \times 10^2 \, \text{ # Example 5.10: Circus Acrobat### GoalUse conservation of mechanical energy to solve a one-dimensional problem involving gravitational potential energy and spring potential energy.### ProblemA 50.0 kg circus acrobat drops from a height of 2.00 meters straight down onto a springboard with a force constant of 8.00 x 10³ N/m, as shown in the figure. By what maximum distance does she compress the spring?### StrategyNonconservative forces are absent, so conservation of mechanical energy can be applied. At the two points of interest, the acrobat's initial position and the point of maximum spring compression, her velocity is zero, so the kinetic energy terms will be zero. Choose y = 0 as the point of maximum compression, so the final gravitational potential energy is zero. This choice also means that the initial position of the acrobat is yᵢ = h + d, where h is the acrobat's initial height above the platform and d is the spring's maximum compression.### Diagram Explanation- **Figure (a)**: Shows an acrobat jumping onto a springboard, causing it to compress.- **Figure (b)**: Shows the acrobat at the initial position before jumping.### Solution**Use conservation of mechanical energy.**\[ (KE + PE_g + PE_s)_i = (KE + PE_g + PE_s)_f \]The only nonzero terms are the initial gravitational potential energy and the final spring potential energy.\[ 0 + mg(h + d) + 0 = 0 + 0 + \frac{1}{2}kd^2 \]\[ mg(h + d) = \frac{1}{2}kd^2 \]Substitute the given quantities and rearrange the equation into standard quadratic form.\[ (50.0 \, \text{kg})(9.80 \, \text{m/s}^2)(2.00\, \text{m} + d) = \frac{1}{2}(8.00 \times 10^3 \, \text{N/m})d^2 \]\[ d^2 - (0.123 \, \text{m}) \, d - 0.245 \, \text{m}^2 = 0 \]Solve with the quadratic formula:\[ d = 0.560 \, \text{m} \

Name: **LEARN MORE****REMARKS** The other solution, \( d = -0.437 \, \text
Uploaded: 2024-03-20T06:57:16.000Z
Channel: Bartleby
Description: **LEARN MORE****REMARKS** The other solution, \( d = -0.437 \, \text{m} \), can be rejected because \( d \) was chosen to be a positive number at the outset. A change in the acrobat's center of mass, say, by crouching as she makes contact with the