Energy solutions

Potential energy depends on configuration: raised objects have gravitational potential energy; separated charges have electric potential energy; a deformed spring has elastic potential energy.

Why: Potential energy belongs to a configuration of interacting objects: height in a gravitational field, charge separation, or spring deformation.

A runner increasing speed increases K = ½mv², which is kinetic energy.

Why: Kinetic energy describes motion. For a fixed mass, increasing the runner's speed increases the squared-speed term in ½mv².

Therefore D describes the kinetic-energy change requested by the question.

Why: D explicitly states a speed increase. The other choices point to configurations used to describe potential energy rather than explicitly specifying a kinetic-energy gain.

Neglect friction and air drag; take the bottom as the gravitational datum.

Why: With friction and drag neglected, no dissipative force removes mechanical energy. Setting bottom height to zero makes its gravitational potential energy vanish.

Conservation of mechanical energy gives ½m(3²)+mg(10)=½mv_f².

Why: The snowboarder starts with both kinetic and gravitational potential energy. At the bottom the gravitational part has become additional kinetic energy.

Cancel m and solve v_f = √(3²+2·9.81·10) = 14.325 m/s.

Why: Mass appears in every energy term and cancels. Add the initial speed squared and the squared-speed gain from the 10 m drop before taking the positive root.

Model an instantaneous launch at the hill bottom, followed by motion up a straight 40° frictionless slope; the coaster must still move at 2 m/s when it reaches 30 m height.

Why: Separate the instantaneous launch from the later uphill coast. The minimum-energy case uses exactly the allowed 2 m/s speed at the required height.

The launch energy is E = mgh+½mv_top² = 300(9.81)(30)+½(300)(2²) = 88,890 J.

Why: The launch must supply both the 30 m gravitational-energy gain and the kinetic energy still required at the top.

Equating this with ½mu² gives launch speed u = √(2²+2gh) = 24.343 m/s.

Why: Immediately after the launch, the supplied energy is kinetic. Rearranging ½mu² = E converts that energy requirement into launch speed.

The downslope acceleration magnitude is gsin40° = 6.306 m/s², so the uphill travel time is t = (u−2)/(gsin40°) = 3.543 s.

Why: A straight frictionless slope gives constant downslope acceleration g sin 40°. Use the drop from u to 2 m/s to find the uphill travel time.

Check the distance: s = 30/sin40° = 46.672 m = ½(u+2)t.

Why: Height is only the vertical part of the slope distance. Compare h/sin 40° with endpoint-average speed times time to check both calculations.

Take ground level as zero potential energy and upward as positive. Initially E_i = mgh = 2.5(9.81)(7.5) = 183.9375 J.

Why: A ground-level potential datum makes the initial rest energy simply mgh. Upward-positive separately fixes the later velocity sign.

The first impact removes 10 J, leaving 173.9375 J for the rebound.

Why: The 10 J impact loss is removed from mechanical energy at the first collision. Only the remainder can power the rebound.

At the rebound peak, mgh_b = 173.9375 J, so h_b = 7.0923 m.

Why: At the rebound peak the box is momentarily at rest. Its remaining mechanical energy is all gravitational potential energy.

With no air drag, the same energy is kinetic just before the second impact: v = −√(2gh_b) = −11.796 m/s.

Why: The return fall has no drag, so peak gravitational energy becomes kinetic just above ground. The negative root sign identifies downward velocity.

The source supplies launch energy and angle but neither ball mass nor a drag model. To compare ranges, first state the idealized assumption of equal common mass m, zero air drag, and equal launch/landing heights.

Why: Launch energy does not fix speed without mass. State the common-mass and ideal-flight assumptions before comparing ranges rather than inventing a numerical mass.

Convert launch kinetic energy to speed using v²=2K/m. Projectile range is R=v²sin(2θ)/g=2Ksin(2θ)/(mg).

Why: Solve K = ½mv² for speed squared, then substitute into the equal-height projectile range. This connects the supplied energies to the requested distances.

Rory: R_R=80sin100°/(mg). Phil: R_P=140sin120°/(mg), so R_P/R_R=1.5389. Under these assumptions Phil goes farther.

Why: With the same mass and g, those factors cancel in the range ratio. The different doubled angles still affect which launch goes farther.

The range difference is ΔR=[140sin120°−80sin100°]/(mg)=(4.3281 kg·m)/m for g=9.81 m/s².

Why: Subtract the two range expressions before inserting g. The common mass remains in the denominator, so this is a mass-dependent distance difference.

A numerical distance difference is not identifiable until a common mass is supplied; if their masses are not equal, even which ball goes farther can change.

Why: An unknown common mass prevents a numerical difference in metres. Different masses would also change the two speeds and can reverse the comparison.

Clarify the diagram: after descending the frictionless 25° ramp, the 8 kg block reaches a horizontal spring on a level frictionless floor; use the text’s k = 22.25 N/m.

Why: The stated horizontal floor separates spring compression from the ramp descent. Using the selected textual k avoids mixing two conflicting source values.

No vertical displacement occurs during compression. At maximum compression x = 0.8 m, all contact kinetic energy is spring energy: K_contact = ½kx² = 7.12 J.

Why: On the level floor, gravitational potential energy does not change. At maximum compression the block is stopped, so its contact kinetic energy has become spring energy.

The contact speed is v = √(2K_contact/m) = √(14.24/8) = 1.334 m/s.

Why: Use K = ½mv² with the energy just found and the 8 kg block mass. The positive square root gives its contact speed.

Starting from rest on the incline, mgLsin25° = 7.12 J; the distance along the ramp before spring contact is L = 0.2147 m.

Why: The frictionless descent starts from rest, so the lost gravitational energy supplies the contact kinetic energy. The vertical drop is L sin 25°, not L.

Check: the corresponding vertical drop is Lsin25° = 0.09072 m.

Why: Multiplying the along-ramp distance by sin 25° recovers the height loss. That value also checks that mg times the drop equals 7.12 J.

Each constant-speed upward lift requires work W_1 = mgh = 15(9.81)(0.65) = 95.6475 J.

Why: Constant upward speed requires a lifting force equal to weight. Aligned upward force and displacement give positive lifting work mgh.

For 873 lifts, total mechanical work is W = 873W_1 = 83,501.2 J.

Why: Each lift covers the same height with the same mass. Multiply one lift's work by the number of upward repetitions counted in the model.

Only 50% of fat’s chemical energy becomes mechanical work, so the chemical energy used is W/0.50 = 167,002 J.

Why: Efficiency is useful mechanical output divided by chemical input. A 50% efficiency therefore needs twice the mechanical work in chemical energy.

At 38 MJ/kg, fat mass is (167,002)/(38×10⁶) = 0.0043948 kg = 4.39 g.

Why: Divide chemical energy by energy per kilogram to find the modeled fuel mass. Convert MJ to J first, then kg to g for the final quantity.

Convert bullet mass to 0.025 kg and rise height to 0.070 m. Combined mass is M = 5.8+0.025 = 5.825 kg.

Why: Use SI units for both the bullet mass and vertical rise. After embedding, the bullet and block move together, so both masses belong in M.

During the short embedding collision, horizontal momentum is conserved: (0.025)u = (5.825)V. Kinetic energy is not conserved during embedding.

Why: Negligible external horizontal impulse preserves momentum only during the short embedding stage. Sticking dissipates kinetic energy, so do not conserve it there.

During the later swing, mechanical energy is conserved: ½MV² = Mg(0.070), so V = √(2g·0.070) = 1.1719 m/s.

Why: Once embedded, the pair swings without losses. Its post-impact kinetic energy becomes gravitational potential energy, fixing the common speed just after impact.

Thus u = (5.825/0.025)(1.1719) = 273.06 m/s.

Why: Insert that post-impact speed into the earlier momentum equation. The block-to-bullet mass ratio explains why the original bullet speed is much larger.

Let x be upward distance from the well bottom, with 0 ≤ x ≤ 8 m. Water mass decreases linearly: m_water(x) = 12−(5/8)x kg.

Why: Choose upward travel from the well bottom as x. Losing 5 kg over 8 m gives the stated constant mass-loss slope with height.

At constant speed with no vertical jet thrust, lifting force balances current weight: F(x) = 9.81[1.5+12−(5/8)x] N.

Why: Constant speed requires zero vertical net force under the no-thrust assumption. The lift balances the weight of the bucket plus only the water still inside.

The force–distance graph is a straight line from (0 m,132.435 N) to (8 m,83.385 N); graph force in newtons against distance in meters.

Why: Because remaining mass falls linearly, the needed lifting force does too. The endpoints and units specify the force–distance graph used for work.

Work is the area of this trapezoid: W = ½(132.435+83.385)(8) = 863.28 J.

Why: Work is force–distance area, not one endpoint force times the whole lift. A straight-line force gives the trapezoid's mean endpoint force times 8 m.

Interpret the 2.5 m/s as the speed at the ramp bottom. On the frictionless rise, ½mv_patch² = ½m(2.5²) − mg(0.03).

Why: The frictionless climb transfers some bottom kinetic energy into gravitational potential energy. Use the 0.03 m height to find the energy left at the patch.

Thus v_patch² = 6.25 − 0.5886 = 5.6614 m²/s² and v_patch = 2.379 m/s.

Why: Solve the reduced energy balance for vpatch², then take its positive root. This is the speed entering friction, not the earlier bottom speed.

On the horizontal patch, friction does work −μ_k mgd. At rest, ½mv_patch² = μ_k mgd.

Why: The horizontal patch has N = mg, so kinetic friction removes energy at the rate μkmg per metre. At stopping it has removed all entry kinetic energy.

The stopping distance is d = 5.6614/[2(0.5)(9.81)] = 0.5771 m, shorter than the 0.6 m patch.

Why: Rearrange for the stopping distance and compare it with 0.6 m. The block runs out of kinetic energy before the patch ends.

It stops on the patch and never reaches its far end; an attempted exit-speed calculation would give a negative squared speed.

Why: The far-end energy would be negative if the friction model were extended that far. That signals an unreachable endpoint, so there is no exit speed.

Potential-energy increase is ΔU=mgΔh=65(9.81)(18)=11,477.7 J.

Why: Near Earth's surface, gravitational potential-energy change depends on vertical rise. Multiply the hiker's mass, g and the 18 m height gain.

Efficiency is useful output/input, so E_chemical=ΔU/0.25=45,910.8 J.

Why: Efficiency compares useful gained potential energy with chemical energy consumed. Divide by 0.25 to account for the energy that does not become height gain.

The remaining 34,433.1 J goes into other energy forms in this model.

Why: Subtract the useful gain from the required chemical input. The balance accounts for all input energy without treating it as vanished.

For a position-dependent force use W=∫F(x)dx, not one endpoint force times distance.

Why: The force grows with position, so its work must add contributions across the interval. The area under F versus x handles the changing force.

W=∫₀³6x dx=[3x²]₀³=27 J.

Why: Integrate 6x from the actual starting x = 0 to x = 3 m. Evaluating both bounds gives the net transferred energy.

Work–energy gives 27=½(2)v², so v=√27=5.196 m/s.

Why: This is net work, so it adds to the sled's kinetic energy. Neglect the nudge's initial kinetic energy as instructed, then solve ½mv²=27 J for the approximate final speed.

Vertical equilibrium gives N=mg−Fsin30°=98.1−25=73.1 N.

Why: The upward pull partly supports the crate. With no vertical acceleration, the floor supplies only the weight remaining after that upward component.

Pulling work is W_pull=Fcos30°·4=173.205 J. Frictional work is W_f=−0.20(73.1)(4)=−58.48 J.

Why: The horizontal pull component acts along displacement and does positive work. Kinetic friction opposes displacement and uses the reduced normal force, giving negative work.

The net work is 114.725 J, which becomes kinetic energy because the crate starts from rest.

Why: Gravity and normal force do no work during the horizontal motion. The two horizontal work terms sum to the kinetic-energy change from rest.

Thus v=√(2·114.725/10)=4.790 m/s.

Why: Solve the final kinetic-energy expression for speed. The positive root reports the magnitude of the crate's motion.

Both gravity and friction oppose uphill motion. At distance d, energy balance is ½mv₀²=mgd sin20°+μ_k mgd cos20°.

Why: During uphill sliding, both downhill gravity and downhill friction remove kinetic energy. A distance d raises the skier by d sin 20° and incurs friction μkmg cos 20°.

Cancel m and solve d=v₀²/[2g(sin20°+μ_k cos20°)].

Why: Factor the common md out of the energy losses and cancel mass. The initial speed squared then fixes the along-slope stopping distance.

Substitute: d=36/[2(9.81)(sin20°+0.15cos20°)]=3.799 m.

Why: Insert the slope angle and coefficient together in the denominator. Both losses contribute, so including friction makes the stop sooner than gravity alone.

Initial spring energy is U_s=½(400)(0.20²)=8 J.

Why: The compressed ideal spring initially stores ½kx². The ball starts from rest, so this is the available mechanical-energy store.

While the ball rises 0.20 m with the spring, gravitational energy increases by (0.50)(9.81)(0.20)=0.981 J.

Why: The ball rises while the spring relaxes, so gravitational potential energy increases before separation. That energy cannot also be counted as kinetic energy.

At spring separation K=8−0.981=7.019 J, so v=√(2K/0.50)=5.299 m/s.

Why: At unloaded length the contact force reaches zero and the ball separates. Subtract the gravitational gain from the released spring energy to find its kinetic energy there.

At the highest point all 8 J is gravitational energy: H=8/(0.50·9.81)=1.631 m above the starting compressed position.

Why: At the final peak, the spring is relaxed and the ball has zero speed. The initial 8 J has become gravitational energy measured above the original compressed position.

At the top the inward force is mg+N=mv_top²/R. At the contact limit N=0, so v_top²=gR.

Why: At the top, both gravity and normal force point inward. The lowest contact-maintaining speed occurs when the normal force just reaches zero, not when speed reaches zero.

The top is 2R above the bottom. Energy conservation gives ½mv_bottom²=½mv_top²+mg(2R).

Why: The cart rises by the loop's diameter, 2R. Frictionless energy conservation adds that potential-energy gain to the kinetic energy it must still have at the top.

Therefore v_bottom²=5gR=98.1 m²/s², giving v_bottom=9.905 m/s.

Why: Substitute the contact-limit top speed squared into the energy equation. The gR needed for contact adds to the 4gR needed for the climb.

Merely having enough energy to reach the top with zero speed would not ensure contact.

Why: Reaching the top geometrically is insufficient: without enough top speed, gravity bends the path more sharply than the track and the cart loses contact earlier.

Each kilogram gains gH=9.81(12)=117.72 J of gravitational potential energy.

Why: Dividing mgh by mass gives the height-gain energy for each kilogram of water. No speed change is included in this model.

Output power is mass rate times energy per kilogram: P_out=3.0(117.72)=353.16 W.

Why: Each second the pump lifts 3 kg through the same height. Multiply energy per kilogram by kilograms per second to obtain output joules per second.

Efficiency=P_out/P_in, so P_in=353.16/0.70=504.51 W.

Why: Only 70% of input power becomes useful hydraulic output. Dividing by efficiency gives the larger electrical power needed.

The vertical drop is h=L(1−cos50°)=1.2(1−cos50°)=0.42865 m.

Why: The bob's downward-vertical projection is L cos 50°. Its difference from the full string length gives the vertical height lost on reaching the bottom.

Tension is perpendicular to the bob’s motion, so it does no work.

Why: A fixed-length string pulls along the radius, while the bob moves along the arc. Perpendicular force and displacement mean tension transfers no energy.

Conservation gives mgh=½mv²; v=√(2gh)=2.900 m/s.

Why: With no drag and no work by tension, the gravitational-energy loss becomes kinetic energy. Mass cancels when solving for bottom speed.

With the floor datum U_i=2(9.81)(5)=98.1 J and U_f=2(9.81)(1)=19.62 J. The kinetic-energy gain is 78.48 J.

Why: With floor height zero, compute both positive potential energies. Their difference, not the initial value alone, becomes kinetic energy during the 4 m fall.

With the shelf datum U_i=0 and U_f=2(9.81)(−4)=−78.48 J. The same decrease supplies 78.48 J of kinetic energy.

Why: Changing zero to the shelf makes the platform's height −4 m. The potential-energy decrease is unchanged even though the individual values differ.

Thus v=√(2·78.48/2)=8.859 m/s downward in either calculation.

Why: Use that common kinetic-energy gain in K = ½mv². Direction comes from the downward fall, while the square root gives speed.

Changing the datum shifts potential energies by a constant; it does not change their difference or the motion.

Why: Adding a constant to U shifts every value equally. Energy differences stay fixed, so choosing a different datum cannot alter the fall speed.

Available gravitational energy is E_g=mgh=(2.0×10⁶)(9.81)(40)=7.848×10⁸ J.

Why: The descending water releases gravitational energy mgh. This is the available input before the plant's conversion losses.

Electrical energy is 0.85E_g=6.6708×10⁸ J. Divide by 3.6×10⁶ J/kWh to obtain 185.3 kWh.

Why: Multiply by overall efficiency to find electrical output energy. Converting joules to kWh uses the energy delivered by 1 kW for 3600 s.

In one hour, average output power is 185.3 kWh/1 h=185.3 kW.

Why: Average power is energy divided by elapsed time. A one-hour duration makes the numerical kW value match the energy's kWh value.

During the sticking collision momentum gives common speed v=1(4)/4=1 m/s.

Why: Treat the two carts as the system only during the short impact. Sticking gives one final velocity, and conserved momentum fixes it.

Initial kinetic energy is 8 J; after collision it is ½(4)(1²)=2 J. Thus 6 J is converted internally.

Why: A sticking collision does not preserve kinetic energy. Comparing the before and after totals identifies the 6 J converted internally.

Only the remaining 2 J compresses the spring: ½(800)x²=2.

Why: The spring acts after the collision. It receives only the 2 J of kinetic energy still present, not the original cart's full 8 J.

Therefore x=√(4/800)=0.07071 m.

Why: Solve the spring-energy equation for the positive compression magnitude. Squaring or taking a root must use the energy remaining after impact.

Gravity does positive work W_g=mgd=0.20(9.81)(5)=9.81 J.

Why: Gravity and downward displacement are aligned. Its positive work is the gravitational-energy loss over the 5 m descent.

Drag does negative work W_drag=−0.60(5)=−3.00 J.

Why: The upward drag force opposes the downward displacement. Their dot product is negative, so drag removes 3 J of mechanical energy.

Net work 6.81 J becomes kinetic energy, so v=√(2·6.81/0.20)=8.252 m/s.

Why: Add both works to obtain net work. Starting from rest makes that work the final kinetic energy; solve it for the impact speed.

Initial kinetic energy is K=½(1200)(20²)=240,000 J.

Why: The car stops completely, so its initial translational kinetic energy is the total amount available to distribute during braking.

Energy absorbed by brakes is Q=0.70K=168,000 J.

Why: Only the specified 70% goes into the brake material. Multiply the initial energy by that fraction before calculating its temperature rise.

Use Q=mcΔT: ΔT=168,000/(12·500)=28 °C.

Why: The heat-capacity model Q = mcΔT relates absorbed energy to temperature change. Use the brake mass and specific heat, not the car mass.

The remaining 30% enters other components or the surroundings in this model.

Why: The remaining fraction must go somewhere too. This model assigns it to other materials or surroundings rather than adding it to the brake temperature calculation.

At constant speed the drive force balances resistance: F_drive=600 N.

Why: Constant speed on a straight level road requires zero net horizontal force. The forward drive therefore matches the 600 N resistance.

Drive power is P=Fv=600(15)=9000 W=9 kW.

Why: Mechanical drive power is its aligned force times velocity. This counts the engine's positive energy transfer, even when total kinetic energy is unchanged.

Ten minutes is 600 s, so E=PΔt=9000(600)=5.4×10⁶ J=1.5 kWh.

Why: Power stays constant because force and speed do. Convert minutes to seconds for joules, or hours for kWh, before multiplying by time.

Net work on the vehicle is zero at constant speed, though the engine transfers energy to overcome resistance.

Why: The resistive force does equal negative work while the drive does positive work. Those cancel in net work but still require the engine to supply energy.

The kinetic-energy increase is ΔK=½(1000)(20²−10²)=150,000 J.

Why: Compute the change in ½mv², not the kinetic energy at either endpoint alone. The motor supplies only that increase in this loss-free model.

Constant power gives Δt=ΔK/P=150,000/30,000=5 s.

Why: A constant power transfers equal energy each second. Divide the required kinetic-energy gain by the 30 000 J/s input to find elapsed time.

Because P=Fv, the starting force is 30,000/10=3000 N and the ending force is 30,000/20=1500 N.

Why: For a drive force along motion, P = Fv. Divide the same power by each endpoint speed to find the different endpoint forces.

Constant power therefore produces a decreasing drive force as speed increases.

Why: As v increases at fixed P, F = P/v decreases. Constant power therefore does not imply constant force or constant acceleration.

At x=1 m, U=4 J and K=16−4=12 J, so speed is √(2K/m)=√24=4.899 m/s.

Why: Total mechanical energy is K + U. At x = 1 m subtract the local potential energy from 16 J, then turn the remaining kinetic energy into speed.

At a turning point K=0, so 4x²=16, giving x=±2 m.

Why: A turning point has zero speed and zero kinetic energy. Setting U equal to the total energy finds both allowed endpoints.

The conservative force is F=−dU/dx=−8x N, so at x=1 m it is −8 N.

Why: A conservative force is minus the slope of its potential-energy graph. At positive x the positive slope gives a negative restoring force.

The required centripetal force is T=mv²/r=0.50(4²)/2=4 N.

Why: The string is the only horizontal inward force. It must supply the centripetal mv²/r needed for the stated constant-speed circle.

At every point tension points radially inward while instantaneous displacement is tangential. Their dot product is zero.

Why: The puck's path is tangential at every instant, while tension points inward. Their perpendicular directions make each small work contribution zero.

Thus ∮T·dr=0 J over the revolution. The force changes momentum direction while kinetic energy remains 4 J.

Why: Adding zero work contributions over the lap gives zero total work. Tension turns the momentum without changing its speed or kinetic energy.

The positive rectangular area is 4(2)=8 J; the negative area is −3(3)=−9 J. Net work is −1 J.

Why: Work is the signed area under force versus position. The later 3 m negative segment outweighs the earlier 2 m positive segment by 1 J.

Initial kinetic energy is ½(2)(4²)=16 J. Final kinetic energy is 15 J.

Why: Use net work as the change in kinetic energy. The −1 J total leaves 15 J from the initial 16 J.

At x=2 m the energy has increased to 24 J; the later −9 J cannot stop the cart before x=5 m.

Why: Check energy before and during the retarding segment. Its 9 J removal is smaller than the 24 J available there, so the cart actually reaches x = 5 m.

Thus v=√(2·15/2)=√15=3.873 m/s, still along +x.

Why: Turn the reachable endpoint's 15 J into speed using the 2 kg mass. Since the cart never stops, its velocity still points along +x.

Convert 45 minutes to 0.75 h. E=PΔt=1.2(0.75)=0.90 kWh.

Why: Use hours with kilowatts to obtain kWh. Multiplying power by the 0.75 h runtime gives the heater's consumed energy.

Since 1 kWh=3.6×10⁶ J, E=0.90(3.6×10⁶)=3.24×10⁶ J.

Why: One kWh is 1000 J/s applied for 3600 s. Multiply by 3.6 million to convert the energy to joules.

Cost is (0.90 kWh)($0.20/kWh)=$0.18.

Why: The tariff charges for each kWh used. Multiplying consumption by dollars per kWh cancels the energy units and gives the cost.

Initial kinetic energy is K=½(2)(3²)=9 J.

Why: The initially moving block carries ½mv² of kinetic energy, while both relaxed springs initially store no spring energy.

For parallel springs, U_total=½(k₁+k₂)x²=½(300)x². Set this equal to 9 J.

Why: Parallel springs share the same compression, so their separate ½kx² energies add. At maximum compression the block stops and all 9 J is stored there.

Thus x=√(18/300)=0.24495 m.

Why: Solve the combined spring-energy equation for x. Take the positive root because the requested compression is a magnitude.

Individual energies are U₁=½(100)(0.06)=3 J and U₂=½(200)(0.06)=6 J, totaling 9 J.

Why: Use the same x² in each spring's energy. The stiffer spring stores twice as much, and their sum checks the initial 9 J budget.