Question

1. use the law of conservation of energy (assume no friction nor air resistance) to determine the kinetic and potential energy at the various marked positions along the roller coaster track below. finally, fill in the bars of the bar charts for positions a, b, c, d, and e.

© the physics classroom, 2009

Explanation:

Step1: Recall Energy Conservation

The law of conservation of energy (no friction/air resistance) states \( KE + PE = \text{Total Energy (TE)} \), which is constant. At position E, \( KE = 40000\,\text{J} \), \( PE = 0\,\text{J} \), so \( TE = 40000 + 0 = 40000\,\text{J} \).

Step2: Position A

At A, \( KE = 0\,\text{J} \). So \( PE_A = TE - KE_A = 40000 - 0 = 40000\,\text{J} \).

Step3: Position B

Given \( KE_B = 25000\,\text{J} \). Then \( PE_B = TE - KE_B = 40000 - 25000 = 15000\,\text{J} \).

Step4: Lowest Point (After B, Let's Call It "Bottom")

At the lowest point, \( PE = 0\,\text{J} \) (since height is minimum). So \( KE_{\text{bottom}} = TE - PE_{\text{bottom}} = 40000 - 0 = 40000\,\text{J} \), \( PE_{\text{bottom}} = 0\,\text{J} \).

Step5: Position C

Assume C is at the same height as B (symmetric? Or check TE). Wait, TE is 40000. If we consider the bar charts, but using TE: At C, since it's a peak (like B? Wait, no, the track: after bottom, it goes up to C, then down. Wait, maybe C is at the same height as B? Wait, no, let's use TE. Wait, maybe C is a peak, so \( PE_C = PE_B = 15000\,\text{J} \)? No, wait, maybe I made a mistake. Wait, at E, \( KE = 40000 \), \( PE = 0 \), so TE is 40000. At A, \( KE = 0 \), so \( PE_A = 40000 \). Then at B: \( KE = 25000 \), so \( PE = 40000 - 25000 = 15000 \). At the bottom (the lowest point, let's say between B and C), \( PE = 0 \), so \( KE = 40000 \). Then at C: let's see the next peak. Wait, maybe C is at a height where \( PE_C =? \). Wait, the next point after C has \( PE = 7500\,\text{J} \) (position D's PE? Wait, the problem says "positions A, B, C, D, and E". Let's re-express:

Wait, the point with \( PE = 7500\,\text{J} \) (let's say D? Wait, the label: "KE=____J, PE=7500J" – that's a point (let's call it D). So for D: \( KE_D = TE - PE_D = 40000 - 7500 = 32500\,\text{J} \).

For E: \( PE_E = 5000\,\text{J} \), so \( KE_E = TE - PE_E = 40000 - 5000 = 35000\,\text{J} \)? Wait, no, the problem says at E, \( KE = 40000\,\text{J} \), \( PE = 0\,\text{J} \). Wait, the original problem: "Finally, KE=40 000J, PE=0J" at the last point (E). So TE is 40000.

Let's list all positions:

• A: \( KE = 0 \), so \( PE = 40000 - 0 = 40000\,\text{J} \)
• B: \( KE = 25000 \), so \( PE = 40000 - 25000 = 15000\,\text{J} \)
• Lowest Point (after B): \( PE = 0 \), so \( KE = 40000 - 0 = 40000\,\text{J} \), \( PE = 0\,\text{J} \)
• C: Let's assume C is at the same height as B (since it's a loop? So \( PE_C = 15000\,\text{J} \), then \( KE_C = 40000 - 15000 = 25000\,\text{J} \) (same as B, symmetric)
• Point with \( PE = 7500\,\text{J} \) (let's say D): \( KE_D = 40000 - 7500 = 32500\,\text{J} \)
• E: \( KE = 40000\,\text{J} \), \( PE = 0\,\text{J} \) (given)
• Point E's Previous (with \( PE = 5000\,\text{J} \)): Wait, the problem has "KE=__J, PE=5000J" (E's previous? No, E is last with \( KE=40000 \), \( PE=0 \). Wait, maybe the "KE=__J, PE=5000J" is a point before E. So \( KE = 40000 - 5000 = 35000\,\text{J} \).

Now, bar charts: For each position, KE and PE bars should add to TE (40000). So:

• A: KE bar height = 0, PE bar height = 40000 (since KE=0, PE=40000)
• B: KE bar height = 25000, PE bar height = 15000
• Lowest Point: KE bar height = 40000, PE bar height = 0
• C: If same as B (loop), KE=25000, PE=15000
• D (PE=7500): KE=32500, PE=7500
• Point with PE=5000: KE=35000, PE=5000
• E: KE=40000, PE=0

Answer:

• Position A: \( KE = 0\,\text{J} \), \( PE = 40000\,\text{J} \)
• Position B: \( KE = 25000\,\text{J} \), \( PE = 15000\,\text{J} \)
• Lowest Point: \( KE = 40000\,\text{J} \), \( PE = 0\,\text{J} \)
• Position C: \( KE = 25000\,\text{J} \), \( PE = 15000\,\text{J} \) (assuming symmetric loop)
• Point with \( PE = 7500\,\text{J} \) (D): \( KE = 32500\,\text{J} \), \( PE = 7500\,\text{J} \)
• Point with \( PE = 5000\,\text{J} \): \( KE = 35000\,\text{J} \), \( PE = 5000\,\text{J} \)
• Position E: \( KE = 40000\,\text{J} \), \( PE = 0\,\text{J} \)

(Bar charts: For each position, draw KE and PE bars such that their heights sum to 40000 units. For example, A: KE bar 0, PE bar 40000; B: KE 25000, PE 15000; etc.)