Work, Power and Energy Lesson Test

Answer: A

Correct answer: Joule. Work is force times the distance moved in the direction of the force, so its unit is a newton multiplied by a meter, named the joule. Because work is energy transferred, energy uses the same unit. The tempting mix-up is the watt: power tells how fast work is done, so 1 watt means 1 joule each second. Horsepower is also a power unit, and the newton measures force alone.

• A (correct): Work and energy share the joule, defined as one newton acting through one meter.
• B: The newton measures force; work needs that force multiplied by a distance.
• C: Horsepower is a power rating of about 746 W, describing a rate of doing work.
• D: The watt is a unit of power, equal to one joule per second, not an amount of work.

Answer: C

Correct answer: 600 J. With the force along the motion, W = Fd.

• A: Adding the force and the distance does not give work; the two must be multiplied.
• B: This is ten times too large, a place-value slip when multiplying 40 by 15.
• C (correct): Force times displacement along the motion gives 40 times 15, which is 600 J.
• D: The number is right but watts measure power; work from force and distance comes out in joules.

Answer: A

Correct answer: 0 J. Work equals force times displacement. Because the carton does not move, d = 0 and the work is 0 J, even though holding it tires the guard's muscles.

• A (correct): The carton has no displacement, so the work is zero no matter how large the force or how long it is held.
• B: This multiplies the force by 120 seconds; time is not a displacement, so no work results.
• C: Dividing the force by the time has no physical meaning here; with no movement the work is zero.
• D: This multiplies the force by the time in minutes, but work uses distance, not time.

Answer: D

Correct answer: 0 J. Only the part of a force along the displacement does work. The holding force points straight up while the bag moves sideways, so the angle is 90°, cos 90° = 0, and the work is 0 J.

• A: This multiplies the upward force by the horizontal distance, but the force is perpendicular to the motion.
• B: Dividing the weight by the distance does not produce work, and the perpendicular force does none anyway.
• C: The force does not oppose the motion; being perpendicular, it does neither positive nor negative work.
• D (correct): The upward force is at 90 degrees to the horizontal motion, and cos 90° is zero, so it does no work.

Answer: C

Correct answer: 50 J. Use KE = ½mv² and square the speed first.

• A: This leaves out the one-half in the kinetic energy formula.
• B: Multiplying mass by speed does not give kinetic energy; the speed must be squared and the result halved.
• C (correct): Half of 4 kg times 5 squared (25) gives 2 × 25 = 50 J.
• D: This uses the speed without squaring it: one-half times 4 times 5.

Answer: D

Correct answer: 300 J. Gravitational potential energy is PE = mgh =

• A: This multiplies mass by height but leaves out g, so it is not in joules.
• B: This multiplies by an extra factor of 10, as if g had been applied twice.
• C: A one-half belongs in kinetic and spring energy, not in mgh.
• D (correct): Mass times g times height gives 15 × 10 × 2 = 300 J.

Answer: B

Correct answer: 300 W. Power is

J ÷ 8 s = 300 W.

• A: The number is right, but power is measured in watts, not joules.
• B (correct): Power is work divided by time: 2,400 ÷ 8 = 300 W.
• C: This multiplies work by time instead of dividing.
• D: This is ten times too small, a division slip.

Answer: D

Correct answer: About 2,984 W. One horsepower is about 746 W, so 4 hp ≈

or roughly 3 kW.

• A: Treating one horsepower as one kilowatt overstates it; one horsepower is about 746 W.
• B: Dividing 4 by 746 goes the wrong direction when converting horsepower to watts.
• C: Dividing 746 by 4 inverts the conversion; horsepower must be multiplied by 746.
• D (correct): Multiplying 4 by about 746 W per horsepower gives roughly 2,984 W.

Answer: C

Correct answer: 75%. Efficiency = output ÷ input ×

The other 200 J warms the jack's moving parts.

• A: This divides input by output; an efficiency above 100% signals an inverted ratio.
• B: This misplaces the decimal when changing 0.75 into a percentage.
• C (correct): Useful output over input is 600 ÷ 800 = 0.75, or 75%.
• D: This is the fraction lost to friction and heat, not the useful fraction.

Answer: C

Correct answer: 6 kWh. Energy used is E = Pt. With power in kilowatts and time in hours,

kWh.

• A: This divides the time by the power instead of multiplying them.
• B: Kilowatts measure power; an amount of energy is expressed in kilowatt-hours.
• C (correct): Energy equals power times time: 2 kW × 3 h = 6 kWh.
• D: Dividing power by time does not give energy; the two must be multiplied.

Answer: A

Correct answer: 200 J. Only the component of the pull along the path does work.

• A (correct): Using W = Fd cos θ gives 50 × 8 × 0.5 = 200 J.
• B: Only a force at 90 degrees does no work; at 60 degrees part of the pull acts along the path.
• C: This uses the full 50 N as if it pointed along the path, ignoring the angle.
• D: This uses sin 60° instead of cos 60°; the cosine gives the component along the motion.

Answer: A

Correct answer: −100 J. Friction acts at 180° to the motion, and cos 180° = −1, so W = −Fd =

The negative sign means friction takes energy away from the box.

• A (correct): Friction points opposite the motion, so its work is −25 × 4 = −100 J.
• B: Friction acts along the line of motion, so it does work; only perpendicular forces do none.
• C: The size is right, but a force opposing the motion removes energy and does negative work.
• D: This divides the force by the distance instead of multiplying them.

Answer: B

Correct answer: 400 J. Push:

Friction:

Net work =

which is the pallet's gain in kinetic energy.

• A: The push is larger than friction, so the net work is positive, not negative.
• B (correct): Push work 1,500 J plus friction work −1,100 J gives a net of 400 J.
• C: This adds the friction work as positive, but friction opposes the motion and its work is negative.
• D: This is the work of the push alone, not the sum of all the forces' work.

Answer: D

Correct answer: 20 J. Because the force grows steadily from zero, the work is the triangle's area under the force-distance graph.

• A: This divides the force by the distance, which does not give an area or a work value.
• B: This uses the full 80 N over the whole distance, treating the triangle as a rectangle.
• C: This takes half of 80 N and divides by 0.5 m instead of multiplying.
• D (correct): The force-distance graph is a triangle, so the work is ½ × 80 × 0.5 = 20 J.

Answer: D

Correct answer: 450 kJ. Kinetic energy is proportional to v². The speed ratio is

so the energy grows by 3² = 9: 9 × 50 kJ = 450 kJ.

• A: This multiplies by 27, cubing the speed ratio instead of squaring it.
• B: There is no reason to multiply by 6; tripling the speed multiplies the energy by 3 squared.
• C: This assumes energy grows in step with speed, but kinetic energy depends on speed squared.
• D (correct): Tripling the speed multiplies kinetic energy by 3², which is 9, and 9 × 50 = 450 kJ.

Answer: B

Correct answer: 6 m/s. From rest, all the net work becomes kinetic energy.

So v² = 36 and

• A: This divides the work by the mass, skipping the one-half and the square root.
• B (correct): Setting 360 = ½ × 20 × v² gives v² = 36, so v = 6 m/s.
• C: This leaves out the one-half, giving v² = 18 and a speed that is too low.
• D: This finds v² correctly but forgets to take the square root.

Answer: C

Correct answer: 15 m. The brakes must remove all the kinetic energy.

Distance =

• A: This multiplies mass by speed instead of finding the kinetic energy.
• B: This leaves out the one-half in the kinetic energy, doubling the distance.
• C (correct): The car's kinetic energy is 90,000 J, and 90,000 ÷ 6,000 = 15 m.
• D: This halves the correct result, applying the one-half twice.

Answer: A

Correct answer: −450 J. For gravity, only the vertical displacement matters, not the route. The box weighs 250 N and rises 1.8 m against gravity, so gravity does

• A (correct): Gravity's work is weight times vertical change, negative because the box rises: −250 × 1.8 = −450 J.
• B: This uses the 7 m path length, but gravity's work depends only on the vertical change.
• C: The size is right, but gravity points down while the box moves up, so its work is negative.
• D: This multiplies the 250 N weight by g again, but a weight in newtons already includes g.

Answer: B

Correct answer: 10 m/s. Potential energy lost equals kinetic energy gained, and mass cancels.

• A: This multiplies g by h without the factor 2 or the square root.
• B (correct): Using v = √(2gh) = √(2 × 10 × 5) = √100 = 10 m/s.
• C: This finds v² = 100 but forgets to take the square root.
• D: This leaves out the factor of 2, taking the square root of 50.

Answer: D

Correct answer: 90 N. By Hooke's law, F = kx = 600 N/m × 0.15 m = 90 N.

• A: This divides the spring constant by the stretch instead of multiplying.
• B: Hooke's law has no one-half; the force is simply k times x.
• C: This is the stored energy in joules, ½kx², not the spring force.
• D (correct): Hooke's law gives F = kx = 600 × 0.15 = 90 N.

Answer: D

Correct answer: 8 J. Square the compression first: 0.2² = 0.04.

• A: This is the spring force kx in newtons, not the stored energy.
• B: This leaves out the one-half in the spring energy formula.
• C: This forgets to square the compression, using ½kx instead of ½kx².
• D (correct): Using ½kx² = 0.5 × 400 × 0.04 = 8 J.

Answer: A

Correct answer: 80 N while pulling 10 m of rope. Effort force = load ÷ MA =

In an ideal machine input work equals output work:

so the rope distance is

• A (correct): Effort is 400 ÷ 5 = 80 N, and the rope moves 5 × 2 = 10 m, so input work equals output work.
• B: This ignores the mechanical advantage for the force, giving five times more input work than output.
• C: This multiplies by the mechanical advantage instead of dividing, reversing the machine's trade.
• D: The force is right, but trading force for distance means the rope moves 5 times the 2 m lift.

Answer: C

Correct answer: 180 N. In an ideal ramp, F × ramp length = weight × height. Work needed is

• A: This inverts the ratio, multiplying by the ramp length and dividing by the height.
• B: This is the force to lift the barrel straight up; the ramp lets a smaller force do the job.
• C (correct): Force times ramp length equals weight times height: 900 × 1.2 ÷ 6 = 180 N.
• D: This is the work needed in joules, 900 × 1.2, not the force.

Answer: B

Correct answer: 90 s. Rearrange

to get

J ÷ 500 W = 90 s.

• A: This divides power by work, inverting the relationship t = W/P.
• B (correct): Time equals work divided by power: 45,000 ÷ 500 = 90 s.
• C: This is ten times too long, a place-value slip in the division.
• D: This is ten times too short, a place-value slip in the division.

Answer: A

Correct answer: 7,500 W. For a steady pull, power equals force times speed: P = Fv = 2,500 N × 3 m/s = 7,500 W, or 7.5 kW.

• A (correct): At steady speed P = Fv = 2,500 × 3 = 7,500 W.
• B: The number is right but joules measure energy; power is in watts.
• C: There is no one-half in P = Fv; this halves the correct value.
• D: This divides force by speed instead of multiplying.

Answer: A

Correct answer: 180 W. Work against gravity: W = mgh =

• A (correct): Work is 60 × 10 × 12 = 7,200 J, and 7,200 ÷ 40 = 180 W.
• B: This uses mass instead of weight, leaving out g.
• C: This is the work in joules; it still has to be divided by the 40 s.
• D: This is ten times too large, a place-value slip in the division.

Answer: B

Correct answer: About 1,885 W. Convert rpm to rad/s, then multiply by torque.

• A: This doubles the correct value, an extra factor of 2 in the conversion.
• B (correct): Converting 600 rpm to about 62.8 rad/s and multiplying by 30 N·m gives about 1,885 W.
• C: This multiplies torque by rpm directly; P = τω needs angular speed in rad/s.
• D: This uses 10 revolutions per second without the factor 2π radians per revolution.

Answer: B

Correct answer: 900,000 J. Convert to watts and seconds: 1.5 kW = 1,500 W and 10 min = 600 s. Then E = Pt =

• A: This keeps the power in kilowatts instead of converting to 1,500 W.
• B (correct): Power 1,500 W times 600 s gives 900,000 J.
• C: This uses 10 for the time without converting minutes to seconds.
• D: This is ten times too small, a place-value slip.

Answer: A

Correct answer: 0 J. The cable does 5,000 × 10 = +50,000 J and gravity does −50,000 J. The net work is 0 J, matching the unchanged speed; the cable's work is stored as potential energy instead.

• A (correct): The cable does +50,000 J and gravity does −50,000 J, so the net work is zero and the speed stays the same.
• B: This is gravity's work alone; the cable's positive work cancels it.
• C: This adds the two works as if both were positive, ignoring gravity's negative sign.
• D: This is the cable's work alone; gravity does an equal amount of negative work.

Answer: C

Correct answer: +320 J. Gravity acts downward along the motion, so it does positive work equal to weight times drop: 80 × 4 = +320 J. The rope does −320 J, so the net work is zero.

• A: This divides the weight by the distance instead of multiplying.
• B: This is the rope's work; gravity acts in the direction of the downward motion, so its work is positive.
• C (correct): Gravity points down along the motion, so its work is positive: 80 × 4 = 320 J.
• D: This is the net work; gravity by itself still does positive work on the descending bag.

Answer: B

Correct answer: 6,000 J. Useful output:

Input = output ÷ efficiency =

The extra 2,400 J becomes heat in the motor and gears.

• A: This multiplies the output by the efficiency, which gives less input than output.
• B (correct): Useful output is 3,600 J; dividing by 0.60 gives 6,000 J of input.
• C: This is the ideal value that ignores the 40% lost to heat and friction.
• D: This adds 60% to the output instead of dividing by 0.60.

Answer: B

Correct answer: 165 N. Potential energy gained:

Work against friction:

Total push work = 990 J, so the push force is

• A: This subtracts friction, but friction opposes the push and must be added.
• B (correct): Total work 750 J + 240 J = 990 J, and 990 ÷ 6 = 165 N.
• C: This covers only the potential energy gain and ignores the energy lost to friction.
• D: This is the total work in joules, not the force; it must be divided by 6 m.

Answer: D

Correct answer: 20 m/s. Start with PE =

J and subtract the 10,000 J loss, leaving 40,000 J of kinetic energy.

So v² = 400 and

• A: This is the ideal speed from all 50,000 J and ignores the energy lost to friction.
• B: This finds v² correctly but forgets to take the square root.
• C: This leaves out the one-half in kinetic energy, giving v² = 200.
• D (correct): Kinetic energy left is 50,000 − 10,000 = 40,000 J, so v² = 400 and v = 20 m/s.

Answer: B

Correct answer: 330 J. Potential energy gained:

Kinetic energy gained: ½ × 8 × 4² = 64 J. Useful output = 264 J, so input =

• A: This accounts for efficiency but counts only the potential energy and leaves out the kinetic energy.
• B (correct): Useful output is 200 J + 64 J = 264 J, and 264 ÷ 0.8 = 330 J.
• C: This is the useful energy delivered to the package, ignoring the 80% efficiency.
• D: This multiplies by the efficiency, which wrongly makes the input smaller than the output.

Answer: A

Correct answer: 50 W. Useful work for each: 50 × 200 N × 1.2 m = 12,000 J. The first worker takes 4 min = 240 s, so

The second delivers only 40 W for the same work.

• A (correct): Each does 50 × 200 × 1.2 = 12,000 J, and 12,000 ÷ 240 s = 50 W.
• B: This divides by 4 minutes without converting the time to seconds.
• C: This is the slower worker's power, using 300 s instead of 240 s.
• D: This adds the two workers' powers instead of finding the first worker's alone.

Answer: C

Correct answer: 20 s. Useful power =

Work needed =

Time =

• A: This uses only the 25% lost to heat as the useful power.
• B: This uses the full 2,000 W input as if the hoist were ideal.
• C (correct): Useful power is 1,500 W and the job needs 30,000 J, so 30,000 ÷ 1,500 = 20 s.
• D: This divides the input by 0.75, making the output larger than the input.

Answer: D

Correct answer: 2 m/s. Spring energy becomes kinetic energy.

So v² = 4 and

• A: This finds v² = 4 but forgets to take the square root.
• B: This leaves out the one-half in kinetic energy, giving v² = 2.
• C: This leaves out the one-half in spring energy, using 12 J instead of 6 J.
• D (correct): The spring stores ½ × 1,200 × 0.01 = 6 J, and 6 = ½ × 3 × v² gives v² = 4, so v = 2 m/s.

Answer: C

Correct answer: 2.4 m. Initial KE = ½ × 5 × 8² = 160 J. At 4 m/s, KE = ½ × 5 × 4² = 40 J. The 120 J difference becomes potential energy:

so h = 2.4 m.

• A: This assumes half the speed means half the maximum height, but energy depends on speed squared.
• B: This is the height where the cart stops completely, not where it slows to 4 m/s.
• C (correct): Kinetic energy falls from 160 J to 40 J; the 120 J lost becomes mgh, so h = 120 ÷ 50 = 2.4 m.
• D: This squares the 4 m/s change in speed instead of taking the change in the squares of the speeds.

Answer: D

Correct answer: 6,000 J. The useful output does not change:

At 50% efficiency, input =

Equivalently, the input scales by

• A: This is the useful output, which stays the same; the input must be larger than it.
• B: This scales the input by 50/75, which wrongly makes a less efficient hoist need less input.
• C: This adds 25% of 4,000 J for the drop in efficiency instead of dividing the output by 0.50.
• D (correct): Useful output is 0.75 × 4,000 = 3,000 J, and 3,000 ÷ 0.50 = 6,000 J.

Answer: C

Correct answer: 1,200 W, with four times the kinetic energy. At steady speed, P = Fv, so the new power is

twice the original 600 W. Kinetic energy depends on v², so doubling the speed gives 2² = 4 times the kinetic energy.

• A: This squares the speed ratio for power too; at constant force, power is proportional to speed.
• B: The power is right, but kinetic energy grows with the square of speed, not in proportion to it.
• C (correct): Power is Fv = 400 × 3 = 1,200 W, and doubling the speed multiplies kinetic energy by 2² = 4.
• D: This is the power at the original 1.5 m/s, not at the new speed.

Answer: C

Correct answer: 240 J. Work is the area under a force-distance graph.

Triangle:

Rectangle:

Total work =

• A: This uses the full 60 N for all 5 m, ignoring that the force was smaller during the first 2 m.
• B: This counts only the flat part of the graph and leaves out the sloped first 2 m.
• C (correct): The triangle gives 60 J and the rectangle gives 180 J, for 240 J in all.
• D: This takes half of 60 N times the whole 5 m, treating the entire graph as a triangle.

Answer: A

Correct answer: 75%. Useful work:

Time: 5 min = 300 s, so useful power =

• A (correct): Useful output is 270,000 J in 300 s, or 900 W, and 900 ÷ 1,200 = 0.75.
• B: This uses mass instead of weight, leaving out g, so the output comes out ten times too small.
• C: This divides input by output; efficiency is output over input and can never exceed 100%.
• D: This is the share of the input lost as heat and friction, not the useful share.

Answer: B

Correct answer: One fourth as much. Since

multiplying the speed by a factor multiplies the energy by that factor squared.

Halving the speed gives

of the original kinetic energy.

• A: This treats kinetic energy as proportional to speed, but speed is squared in the formula.
• B (correct): Kinetic energy depends on speed squared, and (1/2)² = 1/4.
• C: This cubes the speed factor; kinetic energy uses the square of speed, not the cube.
• D: Kinetic energy changes whenever speed changes; only the mass is unchanged.

Answer: D

Correct answer: +7,200 J. Work done by gravity depends only on the vertical displacement, not on the path length.

W = 600 N × 12 m = 7,200 J. It is positive because the force (down) and the vertical motion (down) point the same way.

• A: This multiplies by the 80 m cable length, but gravity's work depends only on the vertical drop.
• B: Gravity points down and the trainee moves down, so its work is positive, not negative.
• C: This adds the cable length and the drop together; only the 12 m vertical change counts.
• D (correct): Gravity's work is weight times vertical drop: 600 × 12 = 7,200 J, positive because the trainee moves downward.

Answer: D

Correct answer: 6 m/s. Push:

Friction:

Net work = 450 J.

By the work-energy theorem, starting from rest:

So

• A: This uses only the 800 J done by the push and ignores the negative work of friction.
• B: This drops the one-half in the kinetic energy formula, solving 450 = 25v².
• C: This stops at v² = 36 and forgets to take the square root.
• D (correct): Net work is 450 J, so 450 = ½ × 25 × v², giving v² = 36 and v = 6 m/s.

Answer: B

Correct answer: 80%. Efficiency compares useful output work with input work.

Output:

Input:

• A: This divides input work by output work; a real machine cannot exceed 100% efficiency.
• B (correct): Output is 1,500 × 0.1 = 150 J and input is 250 × 0.75 = 187.5 J, so 150 ÷ 187.5 = 0.80.
• C: This compares only the two forces, 250 ÷ 1,500, and ignores the distances moved.
• D: This compares only the two distances, 0.1 ÷ 0.75, and ignores the forces.

Answer: C

Correct answer: $27.00. Convert power to kilowatts: 1,500 W = 1.5 kW. Total time: 4

h.

Energy =

kWh, and cost =

• A: This multiplies 1,500 W instead of 1.5 kW, forgetting to convert watts to kilowatts.
• B: This is the cost for only one day of running, not the full 30 days.
• C (correct): Energy is 1.5 kW × 120 h = 180 kWh, and 180 × $0.15 = $27.00.
• D: This slips a decimal place in the final multiplication of 180 kWh by $0.15.

Answer: A

Correct answer: 3 m/s. With no losses, potential energy lost equals kinetic energy gained,

and the mass cancels.

• A (correct): All the lost potential energy becomes kinetic energy, so v = √(2 × 10 × 0.45) = √9 = 3 m/s.
• B: This finds v² = 9 but forgets to take the square root.
• C: This leaves out the factor of 2 and computes √(gh) instead of √(2gh).
• D: This simply multiplies g by h; speed comes from the square root of 2gh.

Answer: B

Correct answer: About 20 hp. At steady speed the engine force equals the resisting force, so

Convert to horsepower: 15,000 ÷ 746 ≈ 20 hp.

• A: This multiplies 15 kW by 0.746 instead of dividing the watts by 746.
• B (correct): Power is 600 × 25 = 15,000 W, and 15,000 ÷ 746 is about 20 hp.
• C: This divides the force by the speed, 600 ÷ 25, which is not a power at all.
• D: This treats 15 kW as if it were 15 hp; a horsepower is about 746 W, not 1,000 W.

Answer: C

Correct answer: 400 J. Potential energy is weight times height, PE = mgh, and mg is the weight.

Since the weight is given as 250 N, PE =

Do not multiply by g a second time.

• A: This multiplies by g = 10 again, but 250 N is already a weight, not a mass.
• B: This divides the weight by the height instead of multiplying.
• C (correct): The weight is already in newtons, so PE = 250 × 1.6 = 400 J.
• D: This divides the weight by 10 as if converting it, which is not needed here.