Unit 3: worked solutions

For the 3 kg cart, K = ½(3)(4²) = 24 J.

Why this step: The squared 4 m/s speed and 3 kg mass give the translational motion energy, with units kg·m²/s².

Set the second energy equal: 24 = ½(12)v_B², giving v_B² = 4 m²/s².

Why this step: Equal energy constrains the product m_Bv_B²; the heavier mass must have a smaller squared speed.

Take the speed magnitude v_B = 2 m/s.

Why this step: The positive square root reports speed, while the equation itself cannot select a direction of travel.

Check: four times the mass and half the speed give factor 4·(½)² = 1 in K.

Why this step: This scaling reproduces equal energy and guards against the incorrect direct-inverse-speed answer of 1 m/s.

Applied work is W_pull = 5·4 = +20 J.

Why this step: The 5 N force points along the 4 m displacement, so its parallel component contributes positive energy transfer.

Opposing-force work is W_opp = −1·4 = −4 J.

Why this step: The opposing force has a 180° angle with displacement, giving a negative contribution rather than reducing the displacement itself.

Net work is W_net = 20 − 4 = 16 J.

Why this step: The work-energy theorem uses the sum of all force work on the object; the applied-force work alone would overestimate the energy gain.

From rest, K_i = 0 and ½(2)v_f² = 16, so v_f = 4 m/s.

Why this step: The object model converts net work into translational kinetic-energy change; the positive root is the requested speed magnitude.

Use x = 0.10 m from relaxed length in U_s = ½kx².

Why this step: The given displacement is the spring deformation relevant to elastic storage, not a shift from a gravity-modified equilibrium.

Calculate U_s = ½(300)(0.10²) = 1.5 J.

Why this step: Squaring the 0.10 m deformation before multiplying gives energy units N·m rather than the force units of kx.

At doubled stretch x = 0.20 m, U_s = ½(300)(0.20²) = 6.0 J.

Why this step: The coefficient k is held fixed; the doubled deformation enters twice through the square.

The energy ratio is 6.0/1.5 = 4.

Why this step: This check confirms a quadratic deformation dependence and rejects the tempting linear doubling to 3 J.

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.

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.

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.

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.

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.

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.

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.

Plot K = ½mvₓ² for the fixed positive mass: an upward parabola with minimum at vₓ = 0.

Why this step: A positive coefficient times a squared component is nonnegative and reaches its smallest value when the component vanishes.

Mark equal heights ½m(2 m/s)² at vₓ = −2 m/s and +2 m/s, and zero at vₓ = 0.

Why this step: The opposite signed components have equal squares, so mirror points about zero must have the same energy.

The symmetry shows that K has no leftward or rightward direction.

Why this step: A directional vector component would reverse sign under reversal; kinetic energy instead retains the same scalar value.

The first 2 m make a triangle of area ½(2 m)(6 N) = 6 J.

Why this step: The parallel force rises linearly from zero, so its average over this interval is 3 N rather than the final 6 N.

The next 1 m make a rectangle of area (1 m)(6 N) = 6 J.

Why this step: Here the force is constant, so every part of the displacement has the same 6 N graph height.

Total work is 6 + 6 = 12 J.

Why this step: The two nonoverlapping intervals both lie above the axis, so their positive work contributions add without cancellation.

List the distinct interacting pairs: AB, AC, and BC.

Why this step: Potential energy belongs to each two-mass interaction; exchanging the names of one pair does not create a second interaction.

Each pair contributes −Gm²/r because both masses equal m and every separation is r.

Why this step: The equilateral geometry supplies the same center separation for all three pairs, with infinity as the stated zero.

Add the three contributions: U_total = −3Gm²/r.

Why this step: Pairwise gravitational energies add once per distinct pair; summing AB and BA separately would double-count the same stored interaction.

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.

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.

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.

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.

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.

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.

Transform velocity: v_relative = 6 − 4 = +2 m/s.

Why this step: The observer moves in the same positive direction, so its motion removes 4 m/s from the cart’s ground-relative component.

Compute K_relative = ½(1)(2²) = 2 J.

Why this step: Use the transformed velocity in the kinetic-energy definition; the mass is unchanged in this ordinary nonrelativistic calculation.

The ground value is 18 J; subtracting ½(1)(4²) = 8 J would give 10 J, which is wrong.

Why this step: The correct square is (6 − 4)² = 36 − 48 + 16, not 36 − 16; energy subtraction misses the velocity cross term.

For either lift from the same starting height to height h above it, gravitational work is W_g = −mgh for constant g.

Why this step: Gravity points downward, and its conservative work depends only on the common upward endpoint change, not intermediate turns.

A longer rough route with constant friction magnitude has W_f = −f_kL, with larger |W_f| when L is larger.

Why this step: Sliding friction opposes every traveled segment, so additional path length removes additional mechanical energy even between the same endpoints.

Equal gravitational work therefore does not imply equal friction work for the two paths.

Why this step: Conservative and nonconservative interactions use different dependencies; substituting endpoint height for friction path length would confuse them.

The initial and final values are U_i = −GMm/R and U_f = −GMm/(2R).

Why this step: The question gives center distances, so they can be substituted directly into the general potential rather than a near-surface mgh approximation.

Subtract final minus initial: ΔU = −GMm/(2R) + GMm/R = +GMm/(2R).

Why this step: The initial negative value is subtracted; keeping that sign is what makes the outward move a positive energy change.

On U versus r, the point at 2R lies higher and closer to zero than the point at R.

Why this step: Moving against attraction increases the system’s potential energy even though both plotted absolute values remain negative.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

Write K_old = ½mv² and K_new = ½m(2v)².

Why this step: Mass is fixed, so the only changed factor in translational kinetic energy is the squared speed.

The ratio is K_new/K_old = 4v²/v² = 4.

Why this step: Squaring the factor of two quadruples energy; a linear factor of two would ignore the speed-squared dependence.

Use work W = Fd cos θ with θ = 90°.

Why this step: The problem gives a particle and a force perpendicular to its displacement, so this angle selects the force component along that displacement.

Since cos 90° = 0, W = 0 J.

Why this step: There is no parallel force component through the displacement, so this interaction transfers no energy to the particle’s kinetic motion.

The force may still change motion direction.

Why this step: A zero work result constrains kinetic-energy change, not acceleration; a perpendicular force can turn the velocity without changing speed.

Moving the zero-height reference to y₀ gives U′ = mg(y − y₀) = U − mgy₀.

Why this step: The same reference shift applies to every position in this fixed gravitational model, so it adds one constant offset.

Then ΔU′ = (U_f − mgy₀) − (U_i − mgy₀) = U_f − U_i = ΔU.

Why this step: The reference offsets cancel when subtracting the two states; the fixed lift’s energy transfer cannot depend on where we label zero.

The individual values change, but the energy difference does not.

Why this step: The interaction and height change stayed physical constants of the lift; changing their labels cannot change the required work.

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.

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.

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.

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.

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.

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.

At the two endpoints, v² is (+3)² = 9 and (−3)² = 9 in m²/s².

Why this step: The question compares endpoint energies at equal speed magnitude; the direction signs disappear on squaring.

With the same mass, both endpoints have K = ½m·9, so the energies are equal.

Why this step: A scalar kinetic energy does not take the velocity’s sign; this says nothing about possible energy changes during the reversal itself.

For one constant-force graph segment, area is F_parallelΔx.

Why this step: The graph’s height is the force component doing work, and its width is the corresponding displacement interval.

Sum the signed segment areas to find W.

Why this step: Successive displacement intervals contribute their own work, and below-axis forces must subtract rather than add positive area.

The units are N·m = J, identifying work rather than impulse or power.

Why this step: Impulse would require a time-axis width, and power would divide energy by time; neither operation is present in this graph.

Write the two relaxed-length deformations as x = +d and x = −d.

Why this step: Equal stretch and compression magnitudes differ by direction, while the spring model uses the same k and relaxed reference.

Both energies are ½k(±d)² = ½kd².

Why this step: Squaring removes the deformation sign, so compressing does not create negative elastic energy.

Choose equal elastic energies.

Why this step: The stored energy depends on deformation magnitude squared rather than on whether the restoring force points left or right.

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.

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.

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.

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.

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.

In one dimension, first find v_relative = v_cart − v_observer using the same axis.

Why this step: K describes motion measured by that observer, so the velocity in its defining equation must belong to that observer’s frame.

Then calculate K_relative = ½m(v_relative)².

Why this step: The entire velocity difference is squared; a velocity offset cannot be applied as a simple kinetic-energy subtraction.

For example, v_cart = 6 m/s and v_observer = 4 m/s give 2 J for a 1 kg cart, not 18 − 8 = 10 J.

Why this step: The mismatch exposes the missing cross term in subtracting separate squares, so it rejects the proposed energy-subtraction shortcut.

On the outward leg, sliding friction points opposite displacement, so W_f,out = −f_kL_out for constant magnitude.

Why this step: The force-displacement angle is 180°, making its work negative even when the outward coordinate is positive.

On the return leg, friction reverses direction with the sliding but still opposes displacement, so W_f,return = −f_kL_return.

Why this step: Changing the path direction also changes friction direction; it does not turn the friction work positive.

Total friction work is −f_k(L_out + L_return), generally nonzero despite zero net displacement.

Why this step: This force is path-dependent; traveled length, not endpoint displacement, counts both negative energy transfers.

At finite r, U = −GMm/r is negative; at 2r it is −GMm/(2r).

Why this step: The specified infinity zero makes a bound finite-separation configuration lower in energy than infinite separation.

The second value is less negative, so U increases as separation grows.

Why this step: Multiplying a positive reciprocal distance by the negative constant reverses the comparison; halving its magnitude moves the value upward.

As r becomes arbitrarily large, U approaches zero from below.

Why this step: The reciprocal term tends toward zero without changing sign, so the curve does not become more negative or stay constant.

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.

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.

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.

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.

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.

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.