Please answer all that apply GOAL Use conservation of mechanical energy tosolve a one-dimensional problem involvinggravitational potential energy and spring potentialenergy.PROBLEM A 50.0 kg circus acrobat drops froma height of 2.00 meters straight down onto aspringboard with a force constant of 8.00 x 103N/m, as in the figure. By what maximum distancedoes she compress the spring?STRATEGY Nonconservative forces are absent,so conservation of mechanical energy can beapplied. At the two points of interest, theacrobat's initial position and the point ofmaximum spring compression, her velocity iszero, so the kinetic energy terms will be zero.(a)(b)Choose y = 0 as the point of maximumAn acrobat drops onto a springboard, causing it tocompress.compression, so the final gravitational potentialenergy is zero. This choice also means that the initial position of the acrobat is y; = h + d, where h is theacrobat's initial height above the platform and d is the spring's maximum compression.SOLUTIONUse conservation of mechanical energy.(1) (KE + PE, + PE,), = (KE + PE, + PE,),%3DThe only nonzero terms are the initial0+ mg(h + d) + 0 = 0 + 0 + ½kd?gravitational potential energy and thefinal spring potential energy.mg (h + d) = ½kd?%3!Substitute the given quantities and(50.0 kg)(9.80 m/s2)(2.00 m + d) = V2(8.00 x 103rearrange the equation into standardN/m)d?quadratic form.d2 - (0.123 m) d - 0.245 m2 = 0Solve with the quadratic formula.d = 0.560 mLEARN MOREREMARKS The other solution, d = -0.437 m, can be rejected because d was chosen to be a positivenumber at the outset. A change in the acrobat's center of mass, say, by crouching as she makes contactwith the springboard, also affects the spring's compression, but that effect was neglected. Shockabsorbers often involve springs, and this example illustrates how they work. The spring action of a shockabsorber turns a dangerous jolt into a smooth deceleration, as excess kinetic energy is converted to springpotential energy.QUESTION Is it possible for the acrobat to rebound to a height greater than her initial height? Explain.(Select all that apply.)O No. The kinetic energy that the acrobat gains on the way down is converted entirely back intopotential energy when she reaches the initial height.Yes. Elastic energy is always present in the spring and can give the acrobat greater height thaninitially.O Yes. The acrobat can provide mechanical energy by pushing herself up while in contact with thespringboard.No. There is no external source of energy to provide the potential energy at a greater height.O Yes. The acrobat can bend her knees while falling and then straighten them as if jumping whenbouncing upward again.

Name: Please answer all that apply GOAL Use conservation of mechanical energ
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Description: Please answer all that apply GOAL Use conservation of mechanical energy tosolve a one-dimensional problem involvinggravitational potential energy and spring potentialenergy.PROBLEM A 50.0 kg circus acrobat drops froma height of 2.00 meters straight d

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GÖÁL Use conservation of mechanical energy to solve a one-dimensional problem involving gravitational potential energy and spring potential energy. PROBLEM A 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 in the figure. By what maximum distance does she compress the spring? STRATEGY Nonconservative 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. (a) (b) Choose y = 0 as the point of maximum An acrobat drops onto a springboard, causing it to compress. compression, so the final gravitational potential energy is zero. This choice also means that the initial position of the acrobat is y; = h + d, where h is the acrobat's initial height above the platform and d is the spring's maximum compression. SOLUTION Use conservation of mechanical energy. (1) (KE + PEg + PE,); = (KE + PE, + PE,)S The only nonzero terms are the initial 0 + mg(h + d) + 0 = 0 +0 + ½kd? gravitational potential energy and the final spring potential energy. mg(h + d) = ½kd? Substitute the given quantities and rearrange the equation into standard quadratic form. (50.0 kg)(9.80 m/s²)(2.00 m + d) = ½(8.00 × 103 N/m)d² d2 - (0.123 m) d - 0.245 m2 = 0 Solve with the quadratic formula. d = 0.560 m LEARN MORE 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 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.) O 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. O Yes. The acrobat can provide mechanical energy by pushing herself up while in contact with the springboard. O No. There is no external source of energy to provide the potential energy at a greater height. O Yes. The acrobat can bend her knees while falling and then straighten them as if jumping when bouncing upward again.