Question

Simone, a 47 kg gymnast, is performing the last tumbling pass of her floor routine. The tumbling pass involves running as fast as possible then jumping into the air to complete a double layout with a twist (this exact maneuver, the double layout with a twist, is not important, but a photo is provided on Canvas to show you what this maneuver is). The instant before she jumps into the air, her center of mass (COM) is 0.90 m above the ground (0.15 m height of the spring floor + 0.75m height of her COM). Her total mechanical energy during the approach is the energy she possesses due to her running velocity and due to the height of her center of mass (COM) above the ground. The total mechanical energy that she possesses the last instant she jumps into the air is completely converted (i.e. equal to) the total mechanical energy she has in the air. The peak height she reaches while in the air is 2.5 m above the ground (2.5 m = spring floor height + jump height). When Simone first made contact with the spring floor during the landing, her COM was again 0.90 m above the ground then decreased as a result of the spring floor maximally compressing during the landing. At this point of maximum compression of the spring floor, her vertical velocity was zero and her COM was 0.85 m above the ground. (a) What is Simone’s total mechanical energy while she is in the air? (b) What was Simone’s velocity the instant that she first jumped into the air? (c) What was the average force exerted on Simone by the floor during the impact? (d) How much work was done on Simone by the spring floor from initial contact with the floor to the point of maximum compression?

Peak height 2.50 m ni COMposition at take-off position at landing COM position at max 0.75 m 0.90 m 0.85 m compression Spring Floor ground0.15 m

Answer 1

(a) At the initial stage, the potential energy of the gymnast is zero and the kinetic energy must be maximum. At the maximum height position, the potential energy is maximum, and the kinetic energy must be zero. However, the total energy must be constant Thus, the total mechanical energy is. Ep = mgh= (47 kg)(9.8 m/s2)(2.50 m) = 1151.5 J to tal (b) From the conservation of energy, total initial energy is equal to total final energy. Thus, EtiE, _ intia 'fina 1 2 mv^+mgh = E,otai 2 (47 kg) v (47 kg)(9.8 m/s2)(0.90 m)(1151.5 J) 5.6 m/s (c) The average force is F ma = From the kinematic equation: 2 = 2as a= 2s Now, the force is (0)-(5.6 m/s) 2(0.15 m-0.85 m -0.75 m -v |FI (47 kg) -14739.2 N14739.2 N 25 The negative sign indicates that this force acts in upward direction (d) From work energy theorem, the work done is equal to change in mechanical energy. Thus W = E-E fial wital 2 mv2 mgh mvo^ +mgh mg (h-h,) 1 (47 kg) (02-(5.6 m/s))+(47 kg)(9.8 m/s2) (0.05 m) =-760 J- -work done by the spring floor The work done by the gymnast is w 760 J- -work done by the gymnast Il