Question

a system contains objects that interact through a field. no energy is transferred into or out of the system. over time, its possible the systems potential energy will increase, decrease, or stay the same.
for each possibility, match the behavior of the systems kinetic energy.

potential energy | kinetic energy
stays the same | increases
increases | stays the same
decreases | decreases

Explanation:

To solve this, we use the law of conservation of mechanical energy (since no energy is transferred in/out, total mechanical energy \( E = KE + PE \) is constant, where \( KE \) = kinetic energy, \( PE \) = potential energy).

Step 1: Potential Energy Stays the Same

If \( PE \) is constant (\( \Delta PE = 0 \)), then \( \Delta KE = \Delta E - \Delta PE = 0 - 0 = 0 \) (because \( E \) is constant). Thus, \( KE \) stays the same.

Step 2: Potential Energy Increases

If \( PE \) increases (\( \Delta PE > 0 \)), then \( \Delta KE = \Delta E - \Delta PE = 0 - \Delta PE < 0 \) (since \( \Delta PE \) is positive). Thus, \( KE \) decreases.

Step 3: Potential Energy Decreases

If \( PE \) decreases (\( \Delta PE < 0 \)), then \( \Delta KE = \Delta E - \Delta PE = 0 - \Delta PE > 0 \) (since \( \Delta PE \) is negative, subtracting a negative is positive). Thus, \( KE \) increases.

Final Matches:

• Potential energy stays the same → Kinetic energy stays the same
• Potential energy increases → Kinetic energy decreases
• Potential energy decreases → Kinetic energy increases

(If the task is to match the columns, the correct pairings are:

• "stays the same" (PE) → "stays the same" (KE)
• "increases" (PE) → "decreases" (KE)
• "decreases" (PE) → "increases" (KE))

Answer:

To solve this, we use the law of conservation of mechanical energy (since no energy is transferred in/out, total mechanical energy \( E = KE + PE \) is constant, where \( KE \) = kinetic energy, \( PE \) = potential energy).

Step 1: Potential Energy Stays the Same

If \( PE \) is constant (\( \Delta PE = 0 \)), then \( \Delta KE = \Delta E - \Delta PE = 0 - 0 = 0 \) (because \( E \) is constant). Thus, \( KE \) stays the same.

Step 2: Potential Energy Increases

If \( PE \) increases (\( \Delta PE > 0 \)), then \( \Delta KE = \Delta E - \Delta PE = 0 - \Delta PE < 0 \) (since \( \Delta PE \) is positive). Thus, \( KE \) decreases.

Step 3: Potential Energy Decreases

If \( PE \) decreases (\( \Delta PE < 0 \)), then \( \Delta KE = \Delta E - \Delta PE = 0 - \Delta PE > 0 \) (since \( \Delta PE \) is negative, subtracting a negative is positive). Thus, \( KE \) increases.

Final Matches:

• Potential energy stays the same → Kinetic energy stays the same
• Potential energy increases → Kinetic energy decreases
• Potential energy decreases → Kinetic energy increases

(If the task is to match the columns, the correct pairings are:

• "stays the same" (PE) → "stays the same" (KE)
• "increases" (PE) → "decreases" (KE)
• "decreases" (PE) → "increases" (KE))