3.4 Conservation of Energy — Worked Solutions
Work through Topic 3.4
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
For a falling ball–Earth system with negligible air resistance, which account is correct?
- Count gravity work and gravitational-energy loss as separate inputs
- Gravitational energy decreases while kinetic energy increases
- The ball creates energy as it falls
Step 1. Choose ball plus Earth: the gravitational interaction lies inside the system boundary.
Why this step? Both participants are included, so gravity is represented by internal potential-energy change rather than an external energy input.
Step 2. With negligible air resistance and no other transfer, ΔK + ΔU_g = 0.
Why this step? The modeled mechanical energy has no dissipative sink or external work, so an increase in one form must match a decrease in the other.
Step 3. During the fall, ΔU_g is negative and ΔK is positive.
Why this step? Their separation in height decreases while motion energy grows; counting gravity work again would record the same transfer twice.
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Question 2
Mechanical energy decreases because of friction. What follows?
- Total energy is not conserved
- Energy can become thermal energy or sound
- Potential energy must become negative
Step 1. Separate mechanical energy K + U from the system’s total energy.
Why this step? The word mechanical restricts the counted forms, so a decrease in that subtotal does not establish a decrease in every energy form.
Step 2. Friction can convert mechanical energy into thermal energy or transfer some as sound.
Why this step? Contact deformations and microscopic motion provide energy destinations omitted by a mechanical-only account.
Step 3. Include those forms and any boundary transfers in the conservation balance.
Why this step? Energy does not disappear when the modeled subtotal falls; the full account must identify the recipient system and changed forms.
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Question 3
A mass compresses a vertical spring while moving downward. Which energy changes must be included?
- Only spring energy
- Both gravitational and spring potential energy
- Only kinetic energy
Step 1. Include mass, spring, and Earth so both spring and gravity interactions are internal.
Why this step? The vertical motion changes two system configurations, not just the deformation of the spring.
Step 2. For downward movement x, gravity contributes ΔU_g = −mgx while spring deformation changes U_s = ½kx².
Why this step? The object loses height while deforming the spring; those potential changes have different signs and dependencies.
Step 3. Both potential terms must be included, along with any kinetic-energy change.
Why this step? Gravity keeps exchanging energy during compression; ignoring it after first contact would predict the wrong turning point.
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Question 4
A 2 kg object starts from rest 1.5 m above a floor. Before reaching the floor, 6 J goes into thermal energy. Find its kinetic energy and speed just before landing using g = 10 m/s².
Step 1. Choose zero gravitational potential at the floor. Initially U_g = mgh = 2·10·1.5 = 30 J and K_i = 0.
Why this step? The object starts from rest at a known height, so all initially available modeled mechanical energy is gravitational.
Step 2. Account for the stated 6 J thermal gain: K_f = 30 − 6 = 24 J just before landing.
Why this step? The energy converted to thermal form is no longer available for the object’s kinetic energy; final gravitational potential is zero at this height.
Step 3. Solve ½mv_f² = 24: v_f = √(2·24/2) = √24 ≈ 4.90 m/s.
Why this step? The kinetic-energy definition determines a speed magnitude, not a velocity sign; taking 30 J here would ignore the specified dissipation.
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Question 5
A cart moves between two states with ΔK = +4 J, ΔU = −7 J, and ΔE_thermal = +1 J. Find the external work and explain its sign.
Step 1. Use W_external = ΔK + ΔU + ΔE_thermal for the stated energy forms and no other transfer.
Why this step? External work must match the whole selected system’s energy change, including the thermal gain rather than only its mechanical subtotal.
Step 2. Substitute W_external = 4 − 7 + 1 = −2 J.
Why this step? The 7 J potential decrease supplies more than the combined 4 J kinetic and 1 J thermal increases, leaving a 2 J outward transfer.
Step 3. The environment removes a net 2 J from the system.
Why this step? With work positive into the system, the negative result identifies energy leaving; it does not mean the thermal energy itself decreased.
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Question 6
Derive maximum compression for a mass m released from rest on a relaxed vertical spring of stiffness k. Compare it with static equilibrium and explain why they differ.
Step 1. Use downward compression x from the relaxed spring length and include mass, Earth, and spring. Initially K = U_s = U_g = 0.
Why this step? The mass starts from rest on the undeformed spring; the gravitational zero is chosen at that starting position for a compact conservation account.
Step 2. At the turning point, K = 0, U_s = ½kx², and U_g = −mgx, so 0 = ½kx² − mgx.
Why this step? The mass is momentarily stopped but has moved downward; gravity’s potential decrease must supply the spring’s stored energy in the lossless model.
Step 3. Factor x(½kx − mg) = 0 and choose the nonzero turning point x_max = 2mg/k.
Why this step? The zero root is the release state; the next stop after compression is the positive nonzero root.
Step 4. Static equilibrium instead satisfies mg − kx_eq = 0, giving x_eq = mg/k.
Why this step? Equilibrium is a force-balance condition, whereas maximum compression is a zero-speed energy condition.
Step 5. The mass reaches equilibrium with nonzero speed and continues to twice that extension before turning.
Why this step? Its accumulated kinetic energy must be stored in additional spring compression; the restoring net force then brings it to rest beyond equilibrium.