Please help me with my WCLN.ca Skate Park- energy graphs 4b questions

onservati this lab, we will c (or moving) Ive friction. In Physics: Energy WCLN.ca Physics: Energy WCLN.ca Energy Graphs (no friction) b. Examine the kinetic energy curve on the graph above. Given that his mass was 75kg This graph was made with a 75 kg Skater riding back and forth on the half-pipe track shown. calculate his speed, v, at the times below using KE = 1/2mv. Without using the simulation, use this graphical data to predict the answers to the following questions. Record your predictions! Speed at zero seconds: show any work below: Energy vs. Time 3.500 Thermal = 0.00 ] Estimate for the kinetic energy , KE , at this time = 3.000 Total = 2918.23 ] KE = 239.56 J 2.500 PE = (2678.67 ) 2.000 - 1 175k / 4mm ) 2 m/s Energy (Joules) 1,500 1.000 Speed at 6.8 seconds: show any work below 11.3 Estimate for the kinetic energy , KE, at this time = J 500 . g. 1 Time (sec ) 17 12 KE - muz 13 8 ( 75ks ) ( 10 . 8 mi ) - v = 10.8 m/s 1 . What types of energy does the "Total Energy" consist of? Kinetic and potential Speed at 8.1 seconds: show any work below What is the magnitude (value) of the total energy throughout? 2 918 .23 J . The grey dotted line on the far right shows what the potential and kinetic energies of the Estimate for the kinetic energy , KE, at this time =] skater are after 1 1.3 seconds on the half pipe. What do these two energies add to? PE + KE = 2918.23 J Ke amv ? Explain why the answer above makes sense . gravitional - 2 ( 75 kg ) ( 12 . 1 mu ) ? V = m/s Because the skater has more, potential energy when his is at the top . As the Skater a Speed at 5.3 seconds: show any work below down it gets converted into kinetic energy gees Estimate for the kinetic energy , KE , at this time =J energy of motion PEAKE. At 11.3 seconds , was the skater closer to the bottom of the pipe or the top ? Explain how amvz you know . The Top. Because at 11. 3 seconds , the potential ( 75 kg ) ( 1 1 3 any ) 2 v = 4683 7 m/s energy is max. and the kinetic energyis minimum 4. Use the above graphs to answer the questions below. a. If his maximum height is 4 m (measured from the bottom of the track), what is his height at the times below (approximate) 7mv 2 30s .J . zero seconds? my = 3 3000 75 X 10- 4 m 280J . 6.8 seconds ? 3 733 my h = PE 2800 410 : 3.733 m m=mass of object ( kg) youJ . 8.1 seconds? 0. 53 M V= velocity of object ( m/s ) 900J. 5.3 seconds ? @1 , am h = Pymg = 400 2021-07-12 75 x10-0.53 m Page 6 of 12 / = PE/mg EK = Kinetic energy Joules) 902 : 1,am 2021-07-12 75 * /0 Page 7 of 11