Springs and Elastic Systems Lesson Test

Answer: C

Correct answer: To the left, back toward the wall. A spring's restoring force always points back toward its natural length. Stretching it to the right makes it pull the block to the left, toward the wall, which is what the minus sign in

expresses.

• A: This is the direction of the displacement; the spring force points opposite to it.
• B: A spring at its natural length exerts no force, but a stretched one resists with a force of kx.
• C (correct): A stretched spring always pulls back toward its natural length, which here means toward the wall.
• D: The spring lies horizontally, so its force acts along its own length, not downward.

Answer: D

Correct answer: 30 N. Use Hooke's law with the deflection in meters.

• A: This uses 20 (centimeters) in place of 0.2 m, a unit slip that makes the answer 100 times too large.
• B: This is half of kx; the one-half factor belongs in the energy formula, not in Hooke's law.
• C: This comes from dividing k by the stretch instead of multiplying them.
• D (correct): Hooke's law gives 150 N/m times 0.2 m, which is 30 N.

Answer: D

Correct answer: An extension spring, loaded by pulling. Tension, or extension, springs are wound with the coils tight together and have hooks or loops so they can be attached and pulled. Compression springs instead have gaps between coils and flat ends so they can be pushed shorter.

• A: A torsion spring resists twisting about its axis; hooks on each end and closed coils point to a spring that is pulled.
• B: A compression spring has open gaps between coils and flat ends so it has room to be squeezed.
• C: A leaf spring is a stack of flat strips that bend, not a coil with hooks.
• D (correct): Closed coils with hooks or loops are the signature of a tension (extension) spring, which is pulled and pulls back.

Answer: C

Correct answer: Newtons per meter (N/m). Rearranging Hooke's law gives

Force is in newtons and deflection in meters, so the spring constant is measured in newtons per meter (N/m).

• A: The newton measures the spring force itself; stiffness is force per unit of deflection.
• B: The joule is the unit of the energy a spring stores, not of its stiffness.
• C (correct): Since k equals force divided by deflection, its unit is newtons per meter.
• D: Force multiplied by distance gives work or torque, not stiffness.

Answer: A

Correct answer: The spring constant k. The slope of a force-extension line is $\Delta F/\Delta x$, which is the spring constant. A steeper line means a stiffer spring, and the area under the line gives the stored energy.

• A (correct): Slope is the change in force divided by the change in extension, which is exactly the stiffness k.
• B: Stored energy is the area under the line, not its slope.
• C: Natural length is a property of the unloaded spring and is not read from the slope of this graph.
• D: The graph describes the spring's response to loads; its own weight does not appear as the slope.

Answer: D

Correct answer: 2 Hz. Frequency is the reciprocal of the period.

• A: This halves the period rather than inverting it.
• B: This shifts the decimal point instead of taking the reciprocal of 0.5 s.
• C: This repeats the period value; frequency is the reciprocal of the period, not the same number.
• D (correct): Frequency is 1 divided by 0.5 s, which is 2 cycles per second.

Answer: B

Correct answer: To turn bouncing motion into heat. The springs carry the load and let the wheels move over bumps. The shock absorbers are dampers: they turn the up-and-down motion into heat in their fluid, so the car settles quickly instead of bouncing repeatedly.

• A: Storing energy is what springs do; shocks remove energy rather than store it.
• B (correct): Shock absorbers are dampers that convert the oscillation's energy into heat in hydraulic fluid so the car stops bouncing.
• C: Supporting the weight is the job of the springs, which also set how far the car sits down.
• D: Ride height under load is set by the springs, so worn or new shocks do not change it.

Answer: D

Correct answer: It stays permanently longer than before. Past the elastic limit, deformation becomes plastic, meaning permanent. The spring takes a set and stays longer than its original length, and it will give less force at a given position from then on.

• A: Overstretching yields the metal in the stretched direction, so it does not end up shorter.
• B: Full recovery happens only for elastic deformation, which ends at the elastic limit.
• C: An overloaded spring is weakened and gives less force at the same position, not more.
• D (correct): Beyond the elastic limit the metal yields, leaving a permanent set so the spring does not fully return.

Answer: D

Correct answer: 400 N/m. Springs that share the same two plates deflect equally and their forces add, so

N/m. A parallel result must be larger than the stiffest single spring.

• A: This is the series result, product over sum, which applies only when the load passes through one spring and then the other.
• B: This is the difference between the two springs, which has no physical meaning for springs sharing a load.
• C: This is the average of the two stiffnesses; parallel springs add their stiffnesses rather than averaging them.
• D (correct): Parallel springs share the same deflection and their forces add, so their stiffnesses add: 150 + 250 = 400 N/m.

Answer: C

Correct answer: As it passes the equilibrium position. Energy swaps between spring energy and kinetic energy during the motion. At the center the spring is at its natural length and stores nothing, so the kinetic energy and the speed are at their maximum.

• A: At release the mass starts from rest, so its speed there is zero.
• B: At the ends the mass momentarily stops and all the energy is stored in the spring, so speed is zero there.
• C (correct): At equilibrium the spring holds no stored energy, so all the energy is kinetic and the speed is greatest.
• D: The mass is still speeding up at this point, because the spring keeps pushing it toward the center.

Answer: A

Correct answer: 0.30 m. In series, each spring carries the full 60 N, so the stretches add.

The same answer comes from

N/m and

• A (correct): The series stiffness is 200 N/m, and 60 N divided by 200 N/m gives 0.30 m.
• B: This treats the springs as parallel, adding stiffnesses to 900 N/m, but a chain is a series connection.
• C: This is only the stretch of the 600 N/m spring; the softer spring stretches even more.
• D: This is only the stretch of the 300 N/m spring; the 600 N/m spring adds its own 0.10 m.

Answer: D

Correct answer: 1.5 in. Pounds are already a force, so no conversion is needed. Two identical springs in parallel give

/in, so the sag is

• A: This treats the springs as if they were in series, which halves the stiffness instead of doubling it.
• B: This uses one spring alone; with two springs sharing the load, the support is twice as stiff.
• C: This divides the stiffness by the load instead of dividing the load by the stiffness.
• D (correct): The parallel stiffness is 30 lb/in, and 45 lb divided by 30 lb/in gives 1.5 in.

Answer: C

Correct answer: 3 m/s. Stored energy:

So

• A: This divides the stored energy by the mass and stops there, without the factor of 2 or the square root.
• B: This is the value of speed squared; the square root was never taken.
• C (correct): The spring stores 2.25 J, and setting this equal to half of 0.5 kg times speed squared gives a speed of 3 m/s.
• D: This drops the one-half from the kinetic energy formula, setting 2.25 J equal to mass times speed squared.

Answer: A

Correct answer: 24 N. Measure deflection from natural length and convert to meters: 31 cm minus 25 cm is 6 cm, or 0.06 m.

• A (correct): The extension is 31 cm minus 25 cm, or 0.06 m, and 400 times 0.06 gives 24 N.
• B: This multiplies by 6 cm without converting to meters, giving an answer 100 times too large.
• C: This uses the natural length of 0.25 m as if it were the deflection.
• D: This uses the full loaded length of 0.31 m instead of the change from natural length.

Answer: D

Correct answer: 0.05 m. At rest the spring force balances the weight, so

The weight is

• A: This treats the 3 kg mass as a 3 N force instead of converting it to weight with mg.
• B: This halves the correct stretch, as if the spring carried only half the weight.
• C: This is a decimal slip, dividing 30 N by 60 instead of 600.
• D (correct): The weight is 3 times 10, or 30 N, and 30 divided by 600 gives 0.05 m.

Answer: C

Correct answer: 1.5 J. Convert 5 cm to 0.05 m, then use the energy formula.

• A: This plugs in 5 instead of 0.05 m, so the squared centimeters make the answer far too large.
• B: 60 is the spring force kx in newtons, not the stored energy.
• C (correct): One-half times 1200 times 0.05 squared gives 600 times 0.0025, which is 1.5 J.
• D: This leaves out the one-half in the energy formula.

Answer: D

Correct answer: 9 times as much. Stored energy is proportional to $x^2$. The deflection ratio is

so the energy ratio is

with no need to know $k$.

• A: This doubles the deflection ratio instead of squaring it.
• B: This would follow from doubling the deflection, but 6 cm is three times 2 cm.
• C: This would be true for the force, which is linear in deflection, but energy depends on the square.
• D (correct): The deflection triples, and energy goes with deflection squared, so it grows by 3 squared, or 9.

Answer: B

Correct answer: 0.6 s. Since

making $k$ four times larger divides the period by

• A: This multiplies by 4, reversing the effect of stiffness and ignoring the square root.
• B (correct): Period goes as the square root of m/k, so 4 times the stiffness halves the period to 0.6 s.
• C: This divides the period by 4, forgetting that the period depends on the square root of k.
• D: This doubles the period, but a stiffer spring makes the motion faster, not slower.

Answer: A

Correct answer: 540 N. Use the total deflection from natural length: 0.01 m of preload plus 0.008 m of extra travel is 0.018 m.

• A (correct): The total compression from natural length is 0.018 m, and 30,000 times 0.018 gives 540 N.
• B: This is half the correct force, as if the average-force idea for work applied to the force itself.
• C: This is only the preload force with the valve closed, before the extra compression.
• D: This counts only the extra 0.008 m and ignores the preload already in the spring.

Answer: B

Correct answer: 400 N/m. A uniform spring acts like many small springs in series. Cutting it in half leaves half the coils, so each piece has

N/m.

• A: Halving k would mean a softer spring, but fewer coils make each piece stiffer.
• B (correct): Each half has half as many coils, so it stretches half as much under the same load and is twice as stiff.
• C: This is the value for both halves placed side by side in parallel, not for a single half.
• D: Stiffness depends on the number of active coils, so cutting the spring does change k.

Answer: A

Correct answer: 3 J. The extra work is the difference between the two stored energies; square each deflection first, then subtract.

• A (correct): The energy difference is one-half times 800 times (0.10 squared minus 0.05 squared), which is 400 times 0.0075, or 3 J.
• B: This squares the 0.05 m change in stretch, which treats the spring as if it started relaxed.
• C: This is the total energy stored at 0.10 m, not the extra work beyond 0.05 m.
• D: This uses the difference of the squares but leaves out the one-half factor.

Answer: A

Correct answer: Still 0.8 s, about a lower center point. The period is

which contains neither g nor orientation. Hanging the cart lowers its equilibrium by $mg/k$, but it oscillates about that point with the same 0.8 s period.

• A (correct): The period depends only on mass and stiffness, so gravity just shifts where the motion is centered.
• B: Gravity is a constant force, so it moves the equilibrium point rather than speeding up the oscillation.
• C: Gravity does stretch the spring and lower the equilibrium point, but it does not change the period.
• D: A hanging mass that is disturbed still oscillates about its new equilibrium position.

Answer: B

Correct answer: Fatigue from millions of load cycles. Even when stresses stay below yield, a spring that is loaded and unloaded over and over can develop tiny cracks that grow with each cycle. This gradual failure is fatigue, and it is the usual way valve and suspension springs break.

• A: The spring was never loaded past yield, so a one-time permanent set is ruled out.
• B (correct): Repeated loading below yield can start small cracks that slowly grow, and this is the most common way valve springs fail.
• C: Damping removes energy as heat; it does not cause cracks to form in the spring wire.
• D: Buckling is a sideways bowing under compression, not a crack that grows over time.

Answer: C

Correct answer: 200 N/m. For springs in series, compliances add directly:

m/N. Stiffness is the reciprocal, so

N/m, which is smaller than either spring as a series result must be.

• A: This is the combined compliance in m/N; it still has to be inverted to get stiffness.
• B: This adds the two stiffnesses, 500 and 333 N/m, which is the rule for parallel springs, not series.
• C (correct): In series the compliances add to 0.005 m/N, and the stiffness is the reciprocal, 1/0.005 = 200 N/m.
• D: This is the softer spring alone; a series pair must be softer still.

Answer: B

Correct answer: Mount the spring over a guide rod. Sideways bowing of a slender compression spring is buckling. It is prevented by guiding the spring over a rod or inside a tube, or by using a shorter, wider spring.

• A: More coils make the spring longer relative to its width, which does nothing to stop sideways bowing.
• B (correct): Guiding the spring along a rod keeps it from bowing sideways, which is the standard fix for buckling.
• C: A longer, equally narrow spring is even more slender and more likely to buckle.
• D: Extra compression adds load to a spring that is already unstable, making the bowing worse.

Answer: B

Correct answer: 15 N. Force is proportional to stretch, so the scale gives

N per cm. The package weighs

• A: This sets up the proportion upside down, as if a larger stretch meant a smaller force.
• B (correct): The spring gives 2.5 N per cm, so 6 cm corresponds to 2.5 times 6 = 15 N.
• C: This uses 0.4 cm per newton as if it were newtons per centimeter, mixing up stiffness and compliance.
• D: This multiplies the test weight by the new stretch without dividing by the original 4 cm.

Answer: A

Correct answer: 0.09 J. Torsion springs follow the same energy pattern as linear springs, with angle in radians.

• A (correct): One-half times 2 times 0.3 squared gives 1 times 0.09, which is 0.09 J.
• B: This squares the angle but leaves out the one-half factor.
• C: This takes one-half of kappa times theta without squaring the angle.
• D: This is the torque, kappa times theta, which is in N·m of torque rather than stored energy.

Answer: C

Correct answer: Critically damped. Too little damping lets the door overshoot and swing; too much makes it creep shut slowly. The fastest return with no overshoot is critical damping, the target for door closers and vehicle suspensions.

• A: An underdamped door would swing past closed and oscillate with shrinking swings.
• B: An overdamped door does not overshoot, but it creeps back slowly rather than as quickly as possible.
• C (correct): Critical damping is the amount that returns a system to rest in the shortest time without overshooting.
• D: With no damping at all, the door would keep swinging back and forth indefinitely.

Answer: D

Correct answer: 5 m/s². First find the spring force, then apply Newton's second law.

• A: This divides k by the mass and leaves out the 0.1 m deflection.
• B: This multiplies the force by the mass instead of dividing by it.
• C: 20 is the spring force in newtons; it still has to be divided by the mass.
• D (correct): The spring force is 200 times 0.1, or 20 N, and 20 N divided by 4 kg gives 5 m/s².

Answer: C

Correct answer: Soft mounts, which add sag. Isolation works when the machine runs much faster than the mount's natural frequency, so soft mounts are chosen. The price is more static sag and larger movement during start-up.

• A: Stiff mounts raise the natural frequency and pass more vibration, and stiffer springs sag less, not more.
• B: Softer springs lower the natural frequency rather than raising it.
• C (correct): Soft mounts lower the natural frequency well below the running speed, isolating vibration at the cost of more static sag and movement.
• D: Stiff mounts limit motion but transmit the vibration into the floor, defeating the goal.

Answer: A

Correct answer: 0.30 m. Parallel pair:

N/m. That group is in series with the 200 N/m spring:

Stretch:

m (0.225 m in the top spring plus 0.075 m in the pair).

• A (correct): The parallel pair acts as 600 N/m, which in series with 200 N/m gives 150 N/m, and 45/150 = 0.30 m.
• B: This treats all three springs as one series chain, but the two 300 N/m springs share the load side by side.
• C: This counts only the stretch of the 200 N/m spring and leaves out the stretch of the parallel pair.
• D: This adds all three stiffnesses as if every spring were in parallel, giving 800 N/m.

Answer: B

Correct answer: 0.72 J. Parallel stiffness is

N/m, so both springs compress

The stiffer spring holds three times the energy of the softer one (0.72 J versus 0.24 J).

• A: This is the total energy in both springs, not the share held by the 900 N/m spring.
• B (correct): Both springs compress 48/1200 = 0.04 m, so the 900 N/m spring stores one-half of 900 times 0.04 squared, which is 0.72 J.
• C: This is the energy in the 300 N/m spring; in parallel the stiffer spring stores more, not less.
• D: This multiplies the spring's final force by its compression without the factor of one-half.

Answer: B

Correct answer: 2.0 m/s. Stored energy:

After the 1 J loss, 4 J becomes kinetic energy.

• A: This sets m times v squared equal to 4 J, dropping the one-half from the kinetic energy formula.
• B (correct): The spring stores 5 J; subtracting the 1 J loss leaves 4 J, and one-half times 2 times v squared equals 4 gives v = 2 m/s.
• C: This drops the one-half from the spring energy, using 10 J, then subtracts the 1 J loss.
• D: This ignores the 1 J loss and puts all 5 J into kinetic energy.

Answer: A

Correct answer: 2.7 J. The extension is measured from natural length:

• A (correct): The extension is 0.06 m, so the energy is one-half of 1500 times 0.06 squared, which is 2.7 J.
• B: This uses the full 26 cm length as the deflection instead of the 6 cm change from natural length.
• C: This plugs in 6 cm as the number 6 without converting it to meters.
• D: This leaves out the one-half in the energy formula.

Answer: B

Correct answer: 2.8 Hz. Two identical springs in series give

Frequency scales as $\sqrt{k/m}$, so it is divided by $\sqrt{2}$.

• A: This doubles the frequency, reversing the effect of the softer series pair and ignoring the square root.
• B (correct): Two identical springs in series have half the stiffness, so the frequency drops by the square root of 2, from 4 Hz to about 2.8 Hz.
• C: This treats the pair as stiffer, as it would be in parallel, but end-to-end springs are softer.
• D: This halves the frequency, but frequency depends on the square root of k, so halving k divides it by the square root of 2.

Answer: D

Correct answer: 35 N. First find the static stretch:

Pulling down 0.03 m more makes the total stretch from natural length 0.07 m.

• A: This adds the 20 N weight on top of the correct spring force, counting the weight twice.
• B: This measures the stretch from the hanging equilibrium point instead of from the spring's natural length.
• C: This is just the weight of the mass, which the spring supports only at the equilibrium position.
• D (correct): The static stretch is 20 N divided by 500, or 0.04 m; adding 0.03 m gives 0.07 m from natural length, and 500 times 0.07 is 35 N.

Answer: B

Correct answer: 7.2 J. For one spring, force is proportional to stretch, and energy is proportional to stretch squared (or

). The force ratio is

so the energy ratio is

• A: This assumes energy is proportional to force, multiplying by 3 instead of 9.
• B (correct): Tripling the force triples the stretch, and energy goes with stretch squared, so it rises by 9 times, to 7.2 J.
• C: This is the extra energy added in going from 40 N to 120 N, not the total stored.
• D: This squares the ratio of the 80 N increase to the original 40 N, which gives 4 instead of 9.

Answer: C

Correct answer: 900 N/m. Each third has one-third as many coils, so its stiffness is

N/m. Two pieces in parallel add:

N/m.

• A: This assumes each piece is one-third as stiff as the original; shorter pieces have fewer coils and are stiffer.
• B: This joins the two 450 N/m pieces in series, but side by side means parallel.
• C (correct): Each third has 3 times the stiffness, 450 N/m, and two in parallel add to 900 N/m.
• D: This simply doubles the original stiffness and ignores the effect of cutting the spring.

Answer: A

Correct answer: 1.5 J. Series stiffness:

N/m, so the force in each spring is

The 600 N/m spring holds 0.75 J, for a total of 2.25 J.

• A (correct): The chain force is 200 N/m times 0.15 m = 30 N, and the softer spring stores 30 squared over 2(300) = 1.5 J.
• B: This is the total energy in the chain, not the 300 N/m spring's share.
• C: This applies the full 0.15 m stretch to the 300 N/m spring alone, but that spring stretches only 0.10 m.
• D: This is the energy in the 600 N/m spring; with equal force in series, the softer spring stores more.

Answer: C

Correct answer: Larger coil diameter and more active coils. For coil springs of the same material, a larger coil diameter and more active coils both lower stiffness, while thicker wire and fewer coils raise it. Only the pair that combines a larger diameter with more coils moves stiffness in the softer direction on both counts.

• A: A larger diameter softens the spring, but fewer coils stiffen it, so the effects partly cancel.
• B: More coils soften the spring, but thicker wire stiffens it, so the two changes work against each other.
• C (correct): A wider coil is softer, and more coils add deflection like springs in series, so both changes reduce stiffness.
• D: Both of these changes make the spring stiffer, the opposite of what is needed.

Answer: A

Correct answer: 4.5 J. The spring force rises from 0 to

so use the average force or the energy formula.

• A (correct): The work equals the stored energy, one half times 400 times 0.15 squared, which is 4.5 J.
• B: This multiplies the final 60 N force by the full 0.15 m, but the force grows from zero, so the average force is only half of 60 N.
• C: This computes one half times k times x without squaring the extension, which does not give an energy.
• D: This is the spring force in newtons at full stretch, k times x, not the work done.

Answer: D

Correct answer: 0.08 m. Simplify from the inside out. The chain:

N/m. The chain and the single spring share the bar and the plate, so they are in parallel:

N/m.

• A: This adds all three stiffnesses as if every spring were in parallel; the 300 and 600 N/m springs are in series with each other.
• B: This treats all three springs as one series chain; the single spring shares the plate with the chain, so it is in parallel.
• C: This puts the 300 and 600 N/m springs in parallel and the 400 N/m spring in series, reversing the real arrangement.
• D (correct): The chain acts as a 200 N/m spring, which shares the plate with the 400 N/m spring, so the total stiffness is 600 N/m and 48 ÷ 600 = 0.08 m.

Answer: D

Correct answer: 18 N. Convert first: 3 cm = 0.03 m. Then

• A: This multiplies 600 by 3 without converting centimeters to meters, making the answer 100 times too large.
• B: This divides k by the deflection; Hooke's law multiplies stiffness by deflection.
• C: This takes half of kx, which belongs in the energy formula, not the force formula.
• D (correct): Converting 3 cm to 0.03 m and multiplying by 600 N/m gives 18 N.

Answer: B

Correct answer: 4 Hz. Frequency is proportional to $\sqrt{k/m}$. The mass became one quarter as large, so the frequency grows by

2 Hz × 2 = 4 Hz.

• A: This scales frequency directly with the mass ratio, forgetting the square root.
• B (correct): The mass dropped by a factor of 4, and frequency varies as the square root of k over m, so the frequency doubles to 4 Hz.
• C: A lighter mass vibrates faster, not slower; this reverses the effect of mass.
• D: This both reverses the direction and skips the square root.

Answer: A

Correct answer: 300 N/m. The spring constant is the slope of the line.

• A (correct): The slope is the change in force over the change in extension: 18 N divided by 0.06 m gives 300 N/m.
• B: This divides 18 N by 6 without converting centimeters to meters; 3 N per cm equals 300 N/m.
• C: This mixes the two points, dividing 30 N by 0.04 m; each force must be paired with its own extension.
• D: This divides the force change by the final 0.10 m extension instead of by the change in extension.

Answer: C

Correct answer: 3.0 m/s. Half a uniform spring is twice as stiff: 900 N/m. Stored energy:

• A: This uses the uncut 450 N/m stiffness for the ball launch; cutting a spring in half doubles its stiffness.
• B: This halves the stiffness to 225 N/m; fewer coils make a spring stiffer, not softer.
• C (correct): Each half is 900 N/m, storing 0.72 J at 0.04 m, and 0.72 J of kinetic energy on the 0.16 kg ball gives 3.0 m/s.
• D: This stops at v squared = 9 and forgets to take the square root.

Answer: B

Correct answer: Underdamped. Underdamped systems swing past equilibrium several times with shrinking amplitude. The ruler loses energy to air drag and internal heating each cycle, so its swings die out.

• A: A critically damped system returns to rest as fast as possible without overshooting, so it would not swing back and forth.
• B (correct): Light damping lets the system overshoot equilibrium repeatedly while the amplitude shrinks, exactly as the ruler does.
• C: An overdamped system creeps slowly back to rest without swinging past equilibrium.
• D: With no damping the swings would keep the same size; here they shrink, so energy is being removed.

Answer: C

Correct answer: 0.20 m. In series every spring carries the full 60 N, so each stretches

Three of them give 0.20 m. Equivalently,

N/m and

• A: This adds the stiffnesses to 2,700 N/m, which is the rule for springs side by side, not end to end.
• B: This is the stretch of just one spring; in series each spring carries the full 60 N and the stretches add.
• C (correct): Three identical springs in series act like one 300 N/m spring, and 60 ÷ 300 = 0.20 m.
• D: This counts only two springs' worth of stretch; all three springs carry the load and each adds its own stretch.

Answer: C

Correct answer: 1.2 N·m. A torsion spring follows the rotational form of Hooke's law:

N·m.

• A: This applies the energy formula, one half κ θ squared, which gives joules of stored energy, not torque.
• B: This takes half of κθ; the half belongs only in the energy formula.
• C (correct): Torque equals angular stiffness times angle: 1.5 × 0.8 = 1.2 N·m.
• D: This divides stiffness by the angle; torque grows with angle, so the two are multiplied.

Answer: D

Correct answer: 4 J. In series both springs carry the same 40 N. The 200 N/m spring stretches

• A: This is the energy in the stiffer 800 N/m spring; with equal force the softer spring stretches more and stores more.
• B: This is the total energy in both springs, not the share in the 200 N/m spring.
• C: This leaves out the one-half in the energy formula, using F squared over k.
• D (correct): In series each spring carries 40 N, so the 200 N/m spring stores 40 squared divided by 400, which is 4 J.