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100 m

(Mon) A roller coaster car loaded with people has a mass of 1000 kg. If the car is raised to the top of the first hill 100 m above the starting point, how much work was done to get the car up here? (5 min/5 pts).

odette-rasmussen
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100 m

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  1. (Mon) A roller coaster car loaded with people has a mass of 1000 kg. If the car is raised to the top of the first hill 100 m above the starting point, how much work was done to get the car up here? (5 min/5 pts)

  2. (Mon) A roller coaster car loaded with people has a mass of 1000 kg. If the car is raised to the top of the first hill 100 m above the starting point, how much work was done to get the car up here? (5 min / 5 pts) W = Fgd W = magd W = (1000 kg)(9.81 m/s²)(100 m) 100 m 9.81 x 105 J

  3. (Tue) A roller coaster car loaded with people has a mass of 1000 kg. If the car is raised to the top of the first hill 100 m above the starting point and is rolling with a speed of 10 m/s, what is its kinetic energy at the top of the hill? (5 min/5pts)

  4. (Tue) A roller coaster car loaded with people has a mass of 1000 kg. If the car is raised to the top of the first hill 100 m above the starting point and is rolling with a speed of 10 m/s, what is its kinetic energy at the top of the hill? (5 min / 5 pts) KE = ½mv² 10 m/s KE = ½(1000 kg)(10 m/s)² KE = 50 kJ 100 m KE = 50 kJ

  5. (Wed) A roller coaster car loaded with people has a mass of 1000 kg. If the car is raised to the top of the first hill 100 m above the starting point and is rolling with a speed of 10 m/s, what is its potential energy at the top of the hill? (5 min/5pts)

  6. (Wed) A roller coaster car loaded with people has a mass of 1000 kg. If the car is raised to the top of the first hill 100 m above the starting point and is rolling with a speed of 10 m/s, what is its potential energy at the top of the hill? (5 min/5 pts) PE = mgh 10 m/s PE = (1000 kg)(9.81 m/s²)(100 m) PE = 981 kJ 100 m KE = 981 kJ

  7. End of Week Procedures • Add up all the points you got this week • Put the total number of points out of 15 points you earnedat the top of your page (no total, lose 5 points) • List any dates you were absent or tardy (and why) • Make sure your name and period # is on the top of the page (no name, no credit) • Turn in your papers

  8. (Wed) A roller coaster car loaded with people has a mass of 1000 kg. If the car at the top of the first hill has an initial ME of 1.031 MJ, and it drops 130 m on the first hill (30 m below the starting point) what is its speed when it reaches the bottom of the first drop (neglect friction)? (5 min/5 pts)

  9. (Wed) A roller coaster car loaded with people has a mass of 1000 kg. If the car at the top of the first hill has an initial ME of 1.031 MJ, and it drops 130 m on the first hill (30 m below the starting point) what is its speed when it reaches the bottom of the first drop (neglect friction)? (5 min/5 pts) MEi = MEf 1.031 MJ = ½mv² + mgh + ½kx² 1.031 MJ 1.031 MJ = ½(1000 kg)v² + (1000 kg)(9.81 m/s²)(-30 m) + 0 130 m 1.031 MJ = 500v² + (-294300 J) (1.031 MJ + 0.2943 MJ) / (500) = v² v² = 2650.6 m²/s² 30 m v = 51.48 m/s

  10. (Thu) A roller coaster car loaded with people has a mass of 1000 kg. Your car with a total ME of 1.031 MJ at the top of the first hill climbs from the first drop to the top of the second hill. If your speed at the top of the second hill is 32.88 m/s, how high above the starting point is the second hill (neglect friction)? (5 min/5 pts)

  11. (Thu) A roller coaster car loaded with people has a mass of 1000 kg. Your car with a total ME of 1.031 MJ at the top of the first hill climbs from the first drop to the top of the second hill. If your speed at the top of the second hill is 32.88 m/s, how high above the starting point is the second hill (neglect friction)? (5 min/5 pts) 1.031 MJ MEi = MEf 1.031 MJ = ½mv² + mgh + ½kx² 1.031 MJ = ½(1000 kg)(32.88 m/s)² + (1000 kg)(9.81 m/s²)(h) 1.031 MJ = 540547.2 J + (9810 kgm/s²)(h) ? m (1.031 MJ) – (540547.2 J) / (9810 kgm/s²) = h h = 50 m

  12. (Fri) Your cell phone slips out of your hand at the top of the second hill on the roller coaster. If you are traveling horizontally at a speed of 32.88 m/s and are 50 m above the ground when you lose it, how far from a point directly under the second hill should you start looking for your cell phone? (10 min/8 pts)

  13. (Fri) Your cell phone slips out of your hand at the top of the second hill on the roller coaster. If you are traveling horizontally at a speed of 32.88 m/s and are 50 m above the ground when you lose it, how far from a point directly under the second hill should you start looking for your cell phone? How long does it take to fall? (y direction) Δx = vit + ½at² ??? 32.88 m/s -50 -50m = 0 + ½(-9.81m/s²)t² 32.88 0 (50/4.905)s² = t² t = 3.19s 0 -9.81 50 m How far does it go horizontally? (x direction) 6.71s 6.71s ??? Δx = vt Δx = (32.88 m/s)(3.19 s) Δx = 104.98 m (8 pts) MEi = MEf ½mvi² + mghi = ½mvf² + mghf If the cell phone has a mass of 100 g, how fast is the cell phone going when it hits the ground? (5 min) ½(0.1)(32.88)² + (0.1)(9.81)(50)= ½(0.1)vf² + 0 (54.05 +49.05)(2)/(0.1)=vf² vf =45.41 m/s (5 pts)

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