Sumo Assignment - ryjrtgdhtds PDF

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SPH4U: Physics, Grade 12, University Preparation Unit 3: Energy and Momentum Activity 4: Elastic Potential Energy

1.What is the gravitational energy (relative to the unstretched surface of the trampoline) of a 20 kg ball at its apex 2.0 m above the trampoline? KE = GPE= mgh mgh = 20 x 10 x 2 Joules= 400 J The kinetic energy just before it hits the trampoline is equal to the gravitational potential energy at its apex. 2. What is the kinetic energy of the ball just before impacting the trampoline? All energy is transferred into kinetic energy therefore Ek = 400J 3.At maximum stretch at the bottom of the motion, what is the sum of the elastic and gravitational energy of the ball? Et3 = Et2 = Et1 = 800J

mgh^3 = mg(-h^2) + 1/2k(-h^2)^2 = 1/2mv^2 = 800J

4.What conclusions can you draw from the answers found above? Through the answers above, I can conclude that energy is always conserved through the law of conservation of energy 5.The sumo wrestler originally only jumps 10 cm above the trampoline but he has the same total energy as the ball which was 200 cm higher than the trampoline. What causes this equivalence?

6.If a 40 kg gymnast and a 400 kg sumo wrestler each dropped from 1 m above the trampoline, find the final position of each athlete. The k value is 12 000 N/m. (Note: This is asking you to find the vertical position of the athlete when the trampoline is at maximum stretch.) potential energy lost = m g h, potential energy gained = (1/2) k x^2

for 40 kg: 40 * 9.81 * (1+x) = (1/2)(12,000) x^2

7.Assume the trampoline is a simple spring that obeys Hooke's law with a k value of 12 000 N/m and the reference height is 0 m when the trampoline is not stretched. When the Sumo is standing on the trampoline (This is the Normal or Contact Plane): a) How far down does the trampoline stretch? Et2 = Ek + Eg + Ee

Et2 = Eg + Ee

Et2 = mg(-{0}2) + 1/2k(-{0}2)^2

b) What is the Total Energy at this location? Et = Ek + Eg + Ee Et1 = Ek

Et1 = 1/2mv^2

Et3 = Ek + Eg + Ee Et3 = Eg

Et2 = Ek + Eg + Ee Et2 = Eg + Ee Et2 = mg(-h2) + 1/2k(-h2)^2

Et3 = mgh^3

Et3 = Et2 = Et1 = 800J

mgh^3 = mg(-h^2) + 1/2k(-h^2)^2 = 1/2mv^2 = 800J [Et=800]

When the Sumo pushes with his legs against the trampoline. The trampoline stretches another 8 cm down. c) What is the Total Energy at this location? Et = Ek + Eg + Ee | Et2 = Eg + Ee | Et2 = mg(-{8}2) + 1/2k(-{8}2)^2 d) Using your results for total energy find the average force exerted by the sumo's leg muscles.

8.Each push of the legs drives the sumo wrestler 10 cm higher in the air. Real world trampolines lose energy since they are damped springs with much internal friction. How much energy does the sumo wrestler lose on each bounce in this situation? The sumo wrestler loses energy equal to the absorbed energy of the trampoline plus the air friction drag loss. With the second question. 9.How can a gymnast keep a constant bounce height in a real world trampoline? By using common factors of physics: weight, gravity, and stability. Weight would keep them at a constant height. Gravity helps the weight & how much force it propels the person, or objects, into the air. Stability helps adjust how much distance the person, or object, needs to be....