3.5 Power — Worked Solutions
Work through Topic 3.5
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
Two useful lifts each raise the same mass through the same height, with equal initial and final speeds and no losses. One takes half the time. Compare average useful output power.
- It is the same
- The faster lift has twice the power
- The faster lift uses half the energy
Step 1. For the same useful lift with equal endpoint speeds and losses treated alike, each transfers gravitational energy W = mgh.
Why this step? Mass and rise fix the gravitational gain; this comparison assumes no unequal kinetic-energy change or additional loss contribution to output work.
Step 2. If the first duration is t and the faster duration is t/2, P_fast = W/(t/2) = 2W/t.
Why this step? Average power divides that same energy amount by elapsed time, so halving only the denominator doubles the rate.
Step 3. The faster lift has twice the average useful lifting power.
Why this step? A shorter duration changes the rate rather than the required mgh; without the stated equal-energy conditions, total motor output power would need more information.
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Question 2
A constant force acts along a cart’s velocity while the cart speeds up. What happens to instantaneous power?
- It increases
- It remains constant
- It becomes zero
Step 1. Because force and velocity are aligned, instantaneous power is P = Fv cos 0° = Fv.
Why this step? Alignment makes the whole force parallel to motion; no angle factor reduces the energy-transfer rate.
Step 2. The fixed F multiplies an increasing speed, so P increases.
Why this step? Equal force over more displacement each second does more work per second; constant force is not a constant-power condition.
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Question 3
What does area under a power–time graph give?
- Energy transferred
- Force
- Acceleration
Step 1. For one constant-power time interval, transferred energy is ΔE = PΔt.
Why this step? Power is energy-transfer rate, so multiplying by the interval undoes the division by time.
Step 2. Add signed interval areas to obtain total energy transferred.
Why this step? A positive-rate segment adds energy under the chosen convention, while a negative-rate segment subtracts it.
Step 3. The area units are W·s = (J/s)·s = J.
Why this step? The time factor cancels the rate denominator; neither force nor acceleration has these resulting units.
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Question 4
A 10 kg object rises 2 m at constant speed in 4 s. Find work, average power, and instantaneous power if the speed is constant throughout, using g = 10 m/s².
Step 1. Constant speed means zero acceleration and equal endpoint kinetic energies, so the upward support force is F = mg = 10·10 = 100 N.
Why this step? No extra net force is needed to change speed, and no dissipation is specified, leaving the support to balance gravity.
Step 2. Work supplied by that lifting force is W = Fh = 100·2 = 200 J.
Why this step? Force and displacement are both upward, so their product is positive and equals the gravitational-energy gain.
Step 3. Average power is W/Δt = 200/4 = 50 W.
Why this step? The 4 s duration is the interval over which the complete 200 J transfer occurs.
Step 4. The constant speed is h/Δt = 2/4 = 0.50 m/s; instantaneous power is Fv = 100·0.50 = 50 W.
Why this step? Both force and speed are constant throughout the modeled lift, so every instantaneous rate equals the interval average.
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Question 5
A cart has 20 J of kinetic energy. A constant 5 W input lasts 6 s with 8 J dissipated. Find the final kinetic energy and identify why input power alone doesn’t give the kinetic-energy increase.
Step 1. Input energy is PΔt = (5 W)(6 s) = 30 J.
Why this step? The stated input power is constant, so its power-time area is a rectangle rather than an unknown variable-rate integral.
Step 2. Balance energy: K_f = K_i + E_input − E_dissipated = 20 + 30 − 8 = 42 J.
Why this step? The initial kinetic energy remains part of the account, and the 8 J diverted to dissipated forms cannot also appear in the final kinetic energy.
Step 3. The kinetic-energy increase is only 22 J, not the full 30 J input.
Why this step? Input power specifies transfer into the system; the internal destination of that energy depends on the separately stated losses.
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Question 6
Derive the slope of v² versus time for a mass m driven at constant power P with no losses. Explain how an experiment could use that slope to infer P.
Step 1. With constant P and no losses, ½mv² − ½mv_i² = Pt.
Why this step? All delivered energy changes translational kinetic energy, and constant power supplies equal energy in equal time intervals.
Step 2. Multiply by 2/m and rearrange: v² = v_i² + (2P/m)t.
Why this step? Isolating v² makes the time dependence linear, even though the speed itself generally does not rise linearly.
Step 3. The v²–t slope is s = 2P/m, so infer P = ms/2.
Why this step? A best-fit slope measures the energy-growth coefficient; multiplying by m/2 converts it back to power units.
Step 4. Measure speeds independently over the constant-power operating interval, square them, and fit against time at known mass.
Why this step? Independent motion data can test the predicted line; values computed from an assumed P would merely redraw the model.
Step 5. Use a range where the drive is actually constant-power and account for friction or rotation if they are significant.
Why this step? Losses or unmodeled rotational energy would make the measured translational slope infer less than the supplied power; the ideal drive need not hold at zero speed.