Statics and Dynamics Quiz: Power And Efficiency

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Power And EfficiencyQuestion 1 of 13

A hydraulic pump delivers fluid at a flow rate of $Q = 0.004 \text{ m}^3/\text{s}$ against a pressure differential of $\Delta p = 3{,}000 \text{ kPa}$ . The mechanical input power to the pump shaft is $18 \text{ kW}$ . Which of the following correctly identifies the pump's hydraulic efficiency and the rate of energy dissipated as heat?

Efficiency

\eta \approx 66.7\%

; heat dissipation rate

\approx 6 \text{ kW}

. The useful hydraulic power is

\Delta p \cdot Q = 3{,}000{,}000 \times 0.004 = 12 \text{ kW}

, and the losses equal the difference between input and output power.

Efficiency

\eta \approx 66.7\%

; heat dissipation rate

\approx 12 \text{ kW}

. The useful hydraulic power is

12 \text{ kW}

, but the heat dissipated equals the output power rather than the power difference.

Efficiency

\eta \approx 150\%

; heat dissipation rate

\approx -6 \text{ kW}

. The useful power exceeds the input because the pressure differential drives fluid work beyond the shaft contribution.

Efficiency

\eta \approx 44.4\%

; heat dissipation rate

\approx 10 \text{ kW}

. The useful hydraulic power is

0.004 / 3{,}000{,}000 = 1.33 \times 10^{-9} \text{ kW}

, so nearly all shaft power is lost.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Power And Efficiency

Practice Power And Efficiency in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Power And Efficiency, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Efficiency $\eta \approx 66.7\%$ ; heat dissipation rate $\approx 6 \text{ kW}$ . The useful hydraulic power is $\Delta p \cdot Q = 3{,}000{,}000 \times 0.004 = 12 \text{ kW}$ , and the losses equal the difference between input and output power. (correct answer)
Efficiency $\eta \approx 66.7\%$ ; heat dissipation rate $\approx 12 \text{ kW}$ . The useful hydraulic power is $12 \text{ kW}$ , but the heat dissipated equals the output power rather than the power difference.
Efficiency $\eta \approx 150\%$ ; heat dissipation rate $\approx -6 \text{ kW}$ . The useful power exceeds the input because the pressure differential drives fluid work beyond the shaft contribution.
Efficiency $\eta \approx 44.4\%$ ; heat dissipation rate $\approx 10 \text{ kW}$ . The useful hydraulic power is $0.004 / 3{,}000{,}000 = 1.33 \times 10^{-9} \text{ kW}$ , so nearly all shaft power is lost.

Explanation: When a question involves pump efficiency, your framework should always be: useful output power ÷ mechanical input power, where useful hydraulic power equals

\Delta p \cdot Q

. Here, the hydraulic output power is

\Delta p \cdot Q = 3{,}000{,}000 \text{ Pa} \times 0.004 \text{ m}^3/\text{s} = 12 \text{ kW}

. Dividing by the shaft input gives efficiency

\eta = 12/18 \approx 66.7\%

. The energy that doesn't reach the fluid must go somewhere — it's dissipated as heat. So the heat loss rate is simply

18 - 12 = 6 \text{ kW}

. That's answer A, which correctly identifies both values. B gets the efficiency right but then claims heat dissipation equals the output power (12 kW) rather than the difference between input and output. This confuses "energy converted to fluid work" with "energy wasted" — a fundamental mix-up of where each power term goes. C inverts the efficiency ratio, dividing input by output instead of output by input, yielding an impossible efficiency above 100%. Real machines always lose some energy; efficiency exceeding 1 violates the first and second laws of thermodynamics. D divides

Q

\Delta p

instead of multiplying — a unit-analysis error that produces a nonsensically tiny power value. Always check that your units cancel correctly:

\text{Pa} \times \text{m}^3/\text{s} = \text{W}

. A reliable study tip: for any energy machine question, write down

P_{\text{out}} = \eta \cdot P_{\text{in}}

and

P_{\text{lost}} = P_{\text{in}} - P_{\text{out}}

before calculating anything. This prevents both the subtraction error in B and the inversion error in C.

Question 2

A 90%-efficient motor drives an 80%-efficient pump. Pump output is 3.6 kW. What is the electrical input?

3.6 kW
4.0 kW
4.5 kW
5.0 kW (correct answer)

Explanation: Work backward from pump output. At 80% efficiency the pump needs 3.6 / 0.8 = 4.5 kW from the motor. The motor is 90% efficient, so electrical input is 4.5 / 0.9 = 5.0 kW. The tempting 4.5 kW ignores the motor's 90% efficiency.

Question 3

An engine of efficiency e drives at constant speed v against drag F. What is the input power?

$F v e$
$Fv/e$ (correct answer)
$Fv$
$F/(ev)$

Explanation: At constant speed, the useful output power equals drag force times velocity, Fv. Efficiency is output divided by input, so input power = output/e = Fv/e. The tempting mistake is Fv alone, which is the output power, not the power you must supply to the engine.

Question 4

A net force accelerates mass m from rest to speed v in time t. What is the average power?

$mv^2/(2t)$ (correct answer)
$mv^2/t$
$mv/t$
$mv^2/2$

Explanation: Average power is work over time. The work done on the mass is its final kinetic energy, mv^2/2, since it starts from rest. Dividing by t gives mv^2/(2t). The tempting mv^2/t forgets the 1/2 factor in the kinetic energy formula.

Question 5

Two cranes lift identical loads to the same height; A takes half as long as B. Which statement must be true?

A's average power is twice B's (correct answer)
A's efficiency is twice B's
A does twice B's useful work
A draws twice B's input power

Explanation: Both cranes do the same useful work because the loads and heights are identical. Since A does that same work in half the time, its average power is double B's. The tempting wrong choice is 'A draws twice B's input power,' but input power depends on efficiency, which isn't given; only average useful power must double.

Question 6

A 75%-efficient pump lifts a 500 kg load 24 m in 40 s. What is the input power? (g=9.8)

2.2 kW
2.9 kW
3.9 kW (correct answer)
5.2 kW

Explanation: Lifting 500 kg 24 m requires 500 x 9.8 x 24 = 117,600 J of work. Over 40 s that's 2,940 W of useful output power. Since the pump is only 75% efficient, divide by 0.75: 2,940 / 0.75 = 3,920 W, or about 3.9 kW. Don't stop at 2.9 kW; that's the output power before accounting for wasted energy.

Question 7

An electric winch motor is rated at $P_{rated} = 5 \text{ kW}$ input electrical power. Its electromechanical efficiency is $\eta_{em} = 80\%$ . The mechanical drive (gears and cable drum) has a separate mechanical efficiency of $\eta_{mech} = 75\%$ . The winch is used to raise a load at constant velocity.

An engineer states: 'The overall efficiency is the average of the two component efficiencies, so $\eta_{overall} = (80\% + 75\%)/2 = 77.5\%$ .' What is the correct overall efficiency, and what maximum load weight can the winch raise at $v = 0.5 \text{ m/s}$ ?

Overall efficiency $= 77.5\%$ ; maximum load weight $= 7{,}750 \text{ N}$ . The engineer's averaging method is correct for efficiencies in series, and the load weight is found from $W = P_{in} \times \eta_{overall} / v$ .
Overall efficiency $= 60\%$ ; maximum load weight $= 6{,}000 \text{ N}$ . The overall efficiency is the product $\eta_1 \times \eta_2 = 0.80 \times 0.75 = 0.60$ , and the load weight is $W = P_{in} \times \eta_{overall} / v = 5{,}000 \times 0.60 / 0.5 = 6{,}000 \text{ N}$ . (correct answer)
Overall efficiency $= 60\%$ ; maximum load weight $= 3{,}000 \text{ N}$ . The overall efficiency is the product $0.80 \times 0.75 = 0.60$ , but the useful power must be divided by velocity squared because power scales with $v^2$ in lifting applications.
Overall efficiency $= 60\%$ ; maximum load weight $= 4{,}000 \text{ N}$ . The overall efficiency is $0.80 \times 0.75 = 0.60$ , and the load weight is $W = P_{in} \times \eta_{overall} / (v \times g)$ where $g = 9.81 \text{ m/s}^2$ must be included as a unit conversion factor.

Explanation: Whenever you see a problem involving multiple efficiency stages in series — like a motor feeding into a gearbox — your first instinct should be to multiply the efficiencies, not average them. Each stage takes the output of the previous one as its input, so losses compound multiplicatively. The correct overall efficiency is

\eta_{overall} = \eta_{em} \times \eta_{mech} = 0.80 \times 0.75 = 0.60

, or 60%. This means only 60% of the electrical input power actually reaches the load. The useful output power is therefore

P_{out} = 5{,}000 \times 0.60 = 3{,}000 \text{ W}

. For constant-velocity lifting, all useful power goes into raising the load, so

P_{out} = W \times v

, giving

W = P_{out}/v = 3{,}000/0.5 = \mathbf{6{,}000 \text{ N}}

. This confirms answer B. A accepts the engineer's averaging claim, which is the central misconception being tested. Averaging efficiencies has no physical basis — it ignores how losses stack in series systems and overstates what power is actually available. C correctly multiplies the efficiencies but then invents a

v^2

scaling rule that doesn't exist. Power for constant-velocity lifting is simply

P = Wv

(linear in

v

, not quadratic), so dividing by

v^2

produces a nonsense result. D correctly finds 60% efficiency but then introduces

g

as a "unit conversion factor," which it is not. Gravitational acceleration is already embedded in the weight

W

(in newtons); there's no reason to divide by

g

again. Study tip: When efficiencies appear in series, always multiply — and remember that

P = Wv

for constant-velocity lifting is a fundamental power relationship worth memorizing cold.

Question 8

A pump operating at steady state draws water from a lower reservoir and discharges it into a pressurized upper tank. The water flow rate is $Q = 0.01 \text{ m}^3/\text{s}$ , the elevation difference is $\Delta z = 10 \text{ m}$ , the pressure in the upper tank is $p_2 = 200 \text{ kPa}$ gauge, and the lower reservoir is open to atmosphere. Neglect velocity head differences and pipe losses. Take $\rho = 1{,}000 \text{ kg/m}^3$ and $g = 10 \text{ m/s}^2$ . The pump's efficiency is $\eta = 70\%$ .

What is the electrical power drawn by the pump motor, assuming the motor itself is 100% efficient (so all electrical power becomes shaft power input to the pump)?

$P_{elec} \approx 1{,}429 \text{ W}$ , found by computing the hydraulic power due to elevation alone as $\rho g Q \Delta z = 1{,}000 \times 10 \times 0.01 \times 10 = 1{,}000 \text{ W}$ and dividing by pump efficiency $\eta = 0.70$ , neglecting the pressure contribution from the upper tank.
$P_{elec} = 3{,}000 \text{ W}$ , found by correctly computing the total hydraulic power as $(\rho g \Delta z + p_2)Q = 3{,}000 \text{ W}$ but then treating this directly as the electrical power drawn, since the motor is stated to be 100% efficient.
$P_{elec} \approx 2{,}857 \text{ W}$ , found by computing only the pressure head contribution as $p_2 Q = 200{,}000 \times 0.01 = 2{,}000 \text{ W}$ and dividing by pump efficiency, on the basis that the elevation term is already embedded in the gauge pressure reference level.
$P_{elec} \approx 4{,}286 \text{ W}$ , found by computing the total hydraulic power including both the elevation head and the gauge pressure in the upper tank as $(\rho g \Delta z + p_2)Q = (100{,}000 + 200{,}000) \times 0.01 = 3{,}000 \text{ W}$ , then dividing by pump efficiency $\eta = 0.70$ . (correct answer)

Explanation: When a pump moves fluid to a higher elevation and into a pressurized vessel, you must account for both contributions to the required hydraulic power. Think of it this way: the pump must overcome the weight of the fluid column (elevation head) and push against the back-pressure in the upper tank. These are independent energy demands that simply add together. The total hydraulic power the pump must deliver to the fluid is:

P_{hydraulic} = (\rho g \Delta z + p_2) \cdot Q = (1{,}000 \times 10 \times 10 + 200{,}000) \times 0.01 = (100{,}000 + 200{,}000) \times 0.01 = 3{,}000 \text{ W}

Since the pump is only 70% efficient, it requires more shaft power than it delivers hydraulically. Because the motor is 100% efficient, all electrical power becomes shaft power, so:

P_{elec} = \frac{P_{hydraulic}}{\eta} = \frac{3{,}000}{0.70} \approx 4{,}286 \text{ W}

That confirms D is correct. Choice A ignores the gauge pressure in the upper tank entirely — a critical omission. The pump must work against that back-pressure regardless of elevation, so dropping it severely underestimates power demand. Choice B correctly computes the total hydraulic power as 3,000 W but then forgets to divide by pump efficiency — the pump's inefficiency means it consumes more power than it delivers to the fluid. Choice C counts only the pressure term and wrongly assumes the elevation head is somehow embedded in the gauge reference — these are physically separate contributions and must both be included explicitly. Study tip: Whenever a pump problem involves both elevation change and a non-atmospheric discharge pressure, write out the full energy equation term by term before plugging in numbers — it's easy to drop a term when you rush straight to a formula.

Question 9

A 2 kg block is pushed along a horizontal surface at constant velocity by a horizontal force. The coefficient of kinetic friction is $\mu_k = 0.3$ and $g = 10 \text{ m/s}^2$ . The block moves at $v = 4 \text{ m/s}$ . If the mechanical efficiency of the pushing mechanism is $\eta = 80\%$ , which of the following correctly describes the input power consumed by the pushing mechanism and the rate of thermal energy generation at the friction interface?

Input power $= 24 \text{ W}$ ; rate of thermal energy generation at the friction interface $= 24 \text{ W}$ . The input power is simply the friction force times velocity, and all of this power is dissipated as heat at the interface; the mechanism efficiency is irrelevant because no kinetic energy is stored at constant velocity.
Input power $= 30 \text{ W}$ ; rate of thermal energy generation at the friction interface $= 30 \text{ W}$ . The input power equals friction power divided by efficiency, and since the block moves at constant velocity all input power — including internal mechanism losses — must ultimately become heat at the sliding interface.
Input power $= 30 \text{ W}$ ; rate of thermal energy generation at the friction interface $= 24 \text{ W}$ . The input power equals the friction power divided by efficiency, and the thermal generation rate at the interface equals the useful output power delivered to the block (friction force times velocity), since all delivered power is dissipated there at constant velocity. (correct answer)
Input power $= 30 \text{ W}$ ; rate of thermal energy generation at the friction interface $= 6 \text{ W}$ . The input power equals friction power divided by efficiency, but only the mechanism's internal losses $(30 - 24 = 6 \text{ W})$ appear as heat, since the 24 W of useful output is entirely consumed in pushing the block and is not converted to thermal energy.

Explanation: When a question combines mechanical efficiency with friction power, you need to track where energy goes at each stage — the mechanism and the sliding interface are two separate dissipation sites. Start by finding the friction force:

f_k = \mu_k mg = 0.3 \times 2 \times 10 = 6 \text{ N}

. The useful output power — the power actually delivered to push the block against friction — is

P_{out} = f_k \times v = 6 \times 4 = 24 \text{ W}

. Since the mechanism is only 80% efficient, it must consume more input power to deliver that 24 W:

P_{in} = \frac{P_{out}}{\eta} = \frac{24}{0.80} = 30 \text{ W}

. The remaining

30 - 24 = 6 \text{ W}

is lost inside the mechanism as heat (gears, linkages, etc.). Now, at the sliding interface, the block moves at constant velocity, so zero kinetic energy accumulates — every watt delivered to the block by the pushing force is immediately consumed by kinetic friction and converted to heat there: 24 W at the interface. This confirms answer C. Answer A ignores efficiency entirely, treating input power as simply

f_k v

, which is actually the output power, not input. Answer B makes a subtler error: it correctly calculates input power as 30 W but then claims all 30 W heats the friction interface, conflating mechanism losses with interface losses — those 6 W are lost inside the mechanism, not at the sliding surface. Answer D inverts the logic, claiming only the mechanism's internal losses appear as heat at the interface, when in reality the delivered 24 W is what becomes heat there. Your strategy: always separate the efficiency stage (mechanism losses) from the output stage (friction dissipation). Efficiency tells you the input cost; friction force times velocity tells you the interface heat rate. They are independent quantities.

Question 10

A machine lifts a 500 N load through a vertical height of 4 m in 5 s. The machine's overall efficiency is 60%. A student claims: 'To find the power drawn from the source, I multiply the useful power output by the efficiency.' Which of the following correctly evaluates this claim and identifies the actual input power?

The student's method is correct. Multiplying useful output power by efficiency yields input power: $P_{in} = P_{out} \times \eta = 400 \times 0.60 = 240 \text{ W}$ .
The student's method is inverted. Input power is found by dividing output power by efficiency: $P_{in} = P_{out}/\eta = 400/0.60 \approx 667 \text{ W}$ , because the source must supply more than the useful output to account for internal losses. (correct answer)
The student's method is inverted, but the correct formula is $P_{in} = P_{out} \times (1 + \eta) = 400 \times 1.60 = 640 \text{ W}$ , because efficiency represents the fractional loss that must be added back to the output power.
The student's method is inverted. However, efficiency applies to the input force rather than the input power, so $P_{in} = (F_{load}/\eta) \times v = (500/0.60) \times (4/5) \approx 667 \text{ W}$ —identical numerically but derived by correcting the applied force rather than the power ratio.

Explanation: Whenever you see a machine efficiency problem, anchor yourself to one core idea: a real machine always wastes some energy, so the source must supply more than the useful output — never less. Efficiency (η) is the ratio of useful output to total input, expressed as

\eta = P_{out}/P_{in}

. Rearranging gives you the key formula:

P_{in} = P_{out}/\eta

. Start by calculating useful output power: the load is 500 N lifted 4 m in 5 s, so

P_{out} = (500 \times 4)/5 = 400 \text{ W}

. Since efficiency is 60%, the input power must be

P_{in} = 400/0.60 \approx 667 \text{ W}

. This makes physical sense — the source supplies 667 W, 400 W does useful work, and the remaining ~267 W is lost to friction and heat. Answer B is correct. Answer A applies the formula backwards. Multiplying output by η gives 240 W, which is less than the useful output — a physically impossible result, since no source can supply less power than the machine actually delivers to the load. Answer C invents a formula with no physical basis. Efficiency does not represent a "fractional loss to add back"; it represents the ratio of output to input. The expression

(1 + \eta)

has no grounding in the definition of efficiency. Answer D arrives at the correct numerical answer (≈667 W) but through faulty reasoning. Efficiency applies to the power ratio, not separately to force. Correcting the force while keeping velocity unchanged is a conceptual workaround that doesn't reflect how efficiency is actually defined. Your go-to memory aid: efficiency less than 1 means input is always larger than output, so you divide output by η to find input — never multiply.

Question 11

A two-stage power transmission system connects a motor to a load. The motor outputs $P_0 = 20 \text{ kW}$ . Stage 1 (a gearbox) has efficiency $\eta_1 = 90\%$ . Stage 2 (a belt drive) has efficiency $\eta_2 = 85\%$ . A technician proposes to improve the overall system by replacing Stage 1 with a higher-efficiency gearbox having $\eta_1' = 95\%$ , keeping Stage 2 unchanged.

By how much does the power delivered to the load increase due to this upgrade?

The power delivered to the load increases by $850 \text{ W}$ , calculated as $20{,}000 \times (0.95 - 0.90) \times 0.85$ , correctly accounting for the fact that Stage 2 acts on the improved Stage 1 output. (correct answer)
The power delivered to the load increases by $1{,}000 \text{ W}$ , calculated as $20{,}000 \times (0.95 - 0.90)$ , because only the efficiency change in Stage 1 matters and Stage 2 does not amplify or attenuate the improvement.
The power delivered to the load increases by $900 \text{ W}$ , calculated as $20{,}000 \times 0.05 \times 0.90$ , because the 5% improvement applies to the original Stage 1 output rather than the motor's full output.
The power delivered to the load increases by $750 \text{ W}$ , calculated as $20{,}000 \times (0.95 \times 0.85 - 0.90 \times 0.85)$ but rounded after applying a correction factor for the interdependence of the two stages, which reduces the net gain.

Explanation: Whenever you see a multi-stage power transmission problem, your first instinct should be to trace power sequentially through each stage. The output of one stage becomes the input to the next — so any change in Stage 1's output gets further scaled by Stage 2's efficiency before reaching the load. The original power delivered to the load is

P = 20{,}000 \times 0.90 \times 0.85 = 15{,}300 \text{ W}

. After the upgrade, it becomes

P' = 20{,}000 \times 0.95 \times 0.85 = 16{,}150 \text{ W}

. The increase is

16{,}150 - 15{,}300 = 850 \text{ W}

, which is exactly

20{,}000 \times (0.95 - 0.90) \times 0.85

. This confirms A is correct: Stage 2 multiplies the improvement from Stage 1 because the extra power gained in Stage 1 must still pass through Stage 2 to reach the load. B is wrong because it ignores Stage 2 entirely, treating the 5% improvement as if it delivers an extra 1,000 W directly to the load. Stage 2 attenuates everything passing through it — including gains. C is wrong because it applies the 5% improvement to the original Stage 1 output (

20{,}000 \times 0.90 = 18{,}000

W) rather than to the motor's full input. The efficiency change applies to the same input power the original stage received: the full motor output. D is wrong because the formula it starts with —

20{,}000 \times (0.95 \times 0.85 - 0.90 \times 0.85)

— actually simplifies to the correct answer of 850 W. The invented "correction factor" is fictional. Strategy tip: Always carry efficiency improvements through every downstream stage. A gain at Stage 1 is still subject to Stage 2's losses — never treat an upstream improvement as arriving at the load unmodified.

Question 12

A vehicle travels at constant speed up a grade inclined at angle $\theta$ to the horizontal. The vehicle mass is $m$ , rolling resistance force is $F_r$ (constant, parallel to road surface), and aerodynamic drag is $F_d = c v^2$ where $c$ is a drag coefficient and $v$ is speed. The engine delivers power $P$ to the drive wheels at overall drivetrain efficiency $\eta$ . Which expression correctly gives the constant speed $v$ the vehicle maintains on this grade?

$v$ satisfies $\eta P = (mg\cos\theta + F_r)v + cv^3$ , because the gravitational component opposing motion along the incline is $mg\cos\theta$ , which is the component perpendicular to the velocity vector on the slope.
$v$ satisfies $P = (mg\sin\theta + F_r + cv^2)v$ , because at constant speed the total resistive force times velocity equals the raw engine power $P$ , with efficiency applied separately as a correction to the left side.
$v$ satisfies $\eta P = (mg\sin\theta + F_r)v + cv^3$ , because the engine's output power must balance the power consumed by gravity, rolling resistance, and aerodynamic drag simultaneously. (correct answer)
$v$ satisfies $\eta P = mg\sin\theta \cdot v + F_r v^2 + cv^3$ , because rolling resistance, like aerodynamic drag, scales with velocity when expressed as a power dissipation term on a slope.

Explanation: When a vehicle moves at constant speed, Newton's second law tells you the net force is zero — meaning the drive force exactly cancels every resistive force. Your job is to translate that force balance into a power balance, then correctly apply drivetrain efficiency. The resistive forces along the road surface are: gravity's component along the slope (

mg\sin\theta

, not cosine — that's the perpendicular component), rolling resistance (

F_r

), and aerodynamic drag (

cv^2

). Power consumed by each force equals force times velocity, so the total resistive power is

(mg\sin\theta + F_r)v + cv^2 \cdot v = (mg\sin\theta + F_r)v + cv^3

. The engine delivers raw power

P

, but drivetrain losses mean only

\eta P

actually reaches the wheels. Setting useful power equal to resistive power gives

\eta P = (mg\sin\theta + F_r)v + cv^3

. That's C, the correct answer. A is wrong because it uses

mg\cos\theta

— the normal force component — instead of

mg\sin\theta

. The component opposing motion along the slope is always the sine term. B makes two errors: it omits

\eta

from the left side (placing efficiency "separately" is physically meaningless here) and writes the drag power term incorrectly as

cv^2 \cdot v

bundled ambiguously inside the parentheses rather than as

cv^3

explicitly. D incorrectly scales rolling resistance with velocity squared (

F_r v^2

). Rolling resistance is modeled as a constant force, so its power contribution is simply

F_r v

, not

F_r v^2

. Study tip: Always build the power equation from scratch — identify each force, multiply by

v

, then apply

\eta

to the delivered side. Watch for sine/cosine confusion on incline problems; draw a free-body diagram if unsure.

Question 13

A motor drives a conveyor belt that lifts packages at a constant velocity. The motor has a rated power output of $12 \text{ kW}$ and operates at an efficiency of $75\%$ . The conveyor lifts packages with a combined weight of $800 \text{ N}$ at a steady rate.

At what maximum constant velocity can the conveyor lift the packages, given the motor's rated power and efficiency?

$v = 15 \text{ m/s}$ , found by dividing the motor's full rated power directly by the load weight without accounting for efficiency losses: $12{,}000 / 800 = 15 \text{ m/s}$ .
$v = 8.44 \text{ m/s}$ , found by dividing the output power by the weight and then dividing again by efficiency, double-counting the efficiency factor: $9{,}000 / (800 \times 1.40) \approx 8.04 \text{ m/s}$ .
$v = 12.5 \text{ m/s}$ , found by multiplying the rated power by efficiency to get an inflated effective power, then dividing by weight: $(12{,}000 \times 1.25) / 800 \times \eta^{-1}$ , incorrectly treating efficiency as a multiplier on the power side.
$v = 11.25 \text{ m/s}$ , found by first computing the useful output power as $P_{out} = \eta \times P_{rated} = 0.75 \times 12{,}000 = 9{,}000 \text{ W}$ , then dividing by the load weight: $v = 9{,}000 / 800 = 11.25 \text{ m/s}$ . (correct answer)

Explanation: Whenever you see a power-and-efficiency problem, your first instinct should be to distinguish between rated (input) power and useful (output) power. A motor's efficiency tells you what fraction of its rated power actually does mechanical work — the rest is lost to heat, friction, and other dissipation. The relationship is simply

P_{out} = \eta \times P_{rated}

. Here, that gives

P_{out} = 0.75 \times 12{,}000 = 9{,}000 \text{ W}

. To find the maximum velocity, recall that power equals force times velocity:

P = Fv

. Since the conveyor lifts at constant velocity, the lifting force equals the package weight, so

v = P_{out}/F = 9{,}000/800 = 11.25 \text{ m/s}

. That's D, the correct answer. A skips efficiency entirely, dividing the full rated power by the weight:

12{,}000/800 = 15 \text{ m/s}

. This treats the motor as 100% efficient, which would violate thermodynamics for any real machine — you never get more useful work out than the rated input. B applies efficiency twice — once to compute output power and again in the denominator — effectively penalizing efficiency twice and arriving at an artificially low velocity. Double-counting a correction factor is a classic trap when the algebra isn't written out carefully. C inverts the efficiency logic, multiplying rated power by a factor greater than 1, as if efficiency somehow amplifies power. Efficiency always reduces available output;

\eta

is always between 0 and 1, never a booster. Study tip: Always write

P_{out} = \eta \times P_{rated}

as your first step in any efficiency problem — it forces you to apply efficiency exactly once before doing anything else.