Work, energy, and power: keep the system account consistent

MCAT content library / Chemistry & physics / Work, energy, and power: keep the system account consistent

A rough strip reduces spring compression without destroying energy. Follow the work on the block and the energy of the larger system to see why both accounts agree.

About 19 minutes, including the worked examples and checks.

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Useful foundations

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Choose what energy change the question asks about
Net work changes kinetic energy
Potential energy packages conservative interactions
Worked application: a ramp, a rough strip, and a spring
Two accounts agree when their boundaries are consistent
Power adds time to the energy account
Check your reasoning
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Choose what energy change the question asks about

A force calculation follows acceleration during motion. An energy calculation can connect two states without first finding every acceleration along the route. The useful account still depends on the chosen object or system and on which interactions transfer or transform energy. AAMC includes work, kinetic and potential energy, conservation, and power in Content Category 4A.¹

For a constant force acting through a straight displacement, W = Fd cos θ, where θ is the angle between the force and displacement. Work is positive for a component along the displacement, negative for an opposing component, and zero for a perpendicular force. Its unit is the joule: 1 J = 1 N·m.²

A block moving horizontally on a fixed level track has no vertical displacement. Its weight and vertical normal force therefore do zero work during that segment. This conclusion uses the specified motion; it is not a rule that normal forces or gravity always do zero work. A force can also be present while doing no work if its point of application does not move.

Net work changes kinetic energy

For the translating block model, the sum of the work done by its forces equals the change in kinetic energy: W_net = ΔK, with K = ½mv². A particular force’s work need not equal ΔK by itself. Positive applied work can be partly or wholly offset by negative work from another force.³

Kinetic energy is a scalar and uses speed squared. Reversing a velocity without changing its magnitude leaves K unchanged. Energy alone therefore does not recover the final direction; a force or motion description supplies that information.

Potential energy packages conservative interactions

Near Earth’s surface, gravitational potential-energy changes are ΔU_g = mgΔh when g is effectively constant. An ideal Hooke’s-law spring stores U_s = ½kx² when displaced by x from its unstretched position. The gravitational reference height can be chosen conveniently; consistent differences determine the prediction.⁴

Gravity and an ideal spring are conservative: their work can be accounted for by a decrease in the corresponding potential energy. If only conservative interactions do work, the mechanical energy K + U remains constant. Sliding friction can instead reduce mechanical energy while increasing internal thermal energy. The larger total-energy account still balances.⁵⁷

Worked application: a ramp, a rough strip, and a spring

All values and diagrams here are original illustrative examples. A 2.0 kg block starts from rest 0.80 m above a horizontal spring. It slides down a smooth ramp, crosses a 2.0 m rough horizontal strip, and reaches a smooth spring-compression region. The strip exerts a constant 3.5 N kinetic friction force. The spring constant is 200 N/m; take g = 10 m/s² and neglect all other losses.

Set the horizontal region’s gravitational potential energy to zero. The starting gravitational energy is (2.0)(10)(0.80) = 16 J. At the bottom of the smooth ramp, that energy is kinetic, so ½(2.0)v² = 16 and v = 4.0 m/s.

Across the rough strip, friction does −(3.5)(2.0) = −7.0 J of work on the moving block. Its kinetic energy falls from 16 to 9.0 J, corresponding to a speed of 3.0 m/s when it reaches the spring. The rough strip ends before spring contact, so friction does not continue during compression.

At maximum compression the block is momentarily at rest. The remaining 9.0 J is spring energy: 9.0 = ½(200)x², giving x = 0.30 m. If the entire route were smooth, 16 J would reach the spring instead, giving x = 0.40 m. These endpoints do not require knowing how long the descent or compression took.

Original energy model for the block, Earth, spring, and track together. Thermal values are increases from the starting state; the table includes the omitted zero entries.

State	K (J)	U_g (J)	U_s (J)	Thermal increase (J)
Start at rest	0	16	0	0
After rough strip	9	0	0	7
Maximum compression, rough route	0	0	9	7
Maximum compression, fully smooth route	0	0	16	0

Original state-by-state energy account

Two accounts agree when their boundaries are consistent

For the block alone, add the work done by each force between the starting rest and maximum compression: gravity contributes +16 J, friction −7 J, and the spring −9 J. The sum is zero, matching ΔK = 0 between those endpoints. The block moved substantially despite zero net work over the full interval.

For the block, Earth, spring, and track together, the same process is a transformation from 16 J of gravitational energy to 9 J of spring energy plus a 7 J thermal increase. Gravity and spring interactions are already represented by potential-energy terms. Adding their work again would count the same contribution twice. Similarly, do not subtract a 7 J friction loss and then separately subtract another 7 J for the thermal rise.

The spring force is not constant during compression. It grows from zero to kx = 60 N. Its average magnitude over this linear force–displacement interval is 30 N, so the work magnitude is (30)(0.30) = 9 J. Multiplying the final 60 N force by the entire 0.30 m would double the correct value. The triangular area under the force–displacement graph explains the factor ½.⁴

Power adds time to the energy account

Average power is energy transferred per elapsed time: P_avg = W/Δt, in watts, where 1 W = 1 J/s. For a specified force, instantaneous mechanical power is F·v. Power can change while the total work required between two specified states stays the same.⁶

An ideal hoist raises a 3.0 kg load by 1.0 m, starting and ending at rest. With g = 10 m/s² and no losses, the hoist delivers 30 J in either a 2.0 s or 6.0 s lift. Its average output powers are 15 W and 5 W. Net work on the load is zero because its endpoint kinetic energies match; the hoist’s positive work is offset by gravity’s negative work. A real motor may require more input energy, depending on efficiency.

Before calculating, name the work or energy being tracked. Include every force’s work when using ΔK for one object, or include the appropriate potential and internal-energy changes for a larger system. Then use the time interval only if the question asks for a transfer rate.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

A block slides right at fixed height along a stationary, smooth horizontal track while compressing an ideal spring. The only forces on the block are its weight, the track’s normal force, and the spring force. Which work statement is correct during compression?

A. All three forces do zero work because the block is at rest at maximum compression.

Review this choice. The block moves during compression. Endpoint rest does not make each force’s work zero; the spring removes the block’s initial kinetic energy.

B. The normal force does positive work that cancels the negative work of gravity.

Review this choice. Both vertical forces are perpendicular to every horizontal displacement. Each does zero work individually; they are not two nonzero work contributions that cancel.

C. Weight and normal force do zero work, while the spring does negative work on the block.

Correct. The vertical forces have no component along the displacement. The spring points left while the block moves right, so its work on the block is negative and the block’s kinetic energy decreases.

D. Weight and normal force do zero work, while the spring does positive work on the block.

Review this choice. The spring’s stored energy increases, but its work on the block is negative. These describe opposite sides of the same energy transfer.

The principle to carry forward

The sign of work belongs to a force acting on a specified object, not simply to whether energy is stored somewhere in the system.

Check 2

Repeat the worked 2.0 kg descent from 0.80 m with g = 10 m/s² and the same 7.0 J friction loss. The rough strip again ends before spring contact, but the spring constant is now 800 N/m. What is the maximum compression?

A. 0.075 m

Review this choice. The spring is four times stiffer, but compression does not decrease fourfold at fixed stored energy. Energy depends on x².

B. 0.15 m

Correct. The spring receives 16 − 7 = 9 J. Thus x² = 2(9)/800 = 0.0225 m² and x = 0.15 m. A fourfold increase in k halves x at fixed energy.

C. 0.30 m

Review this choice. This preserves the old compression despite the increased spring constant. At 0.30 m the new spring would store 36 J, more than the available energy.

D. 0.60 m

Review this choice. A stiffer spring requires less compression to store the same energy. This value reverses the dependence and would require far more than 9 J.

The principle to carry forward

First determine the energy that reaches the changed component. Then apply its energy–configuration relation; do not scale a squared quantity linearly.

Check 3

Two ideal hoists each raise a 5.0 kg load through 0.80 m, starting and ending at rest. One lift takes 2.0 s and the other 5.0 s. Use g = 10 m/s² and neglect losses. Which row gives the work done by each hoist, followed by the average output power for the 2.0 s and 5.0 s lifts?

A. 0 J per lift; 0 W and 0 W.

Review this choice. The endpoint kinetic-energy change is zero, but the question asks for hoist work, not net work on the load. Each hoist must supply the 40 J gravitational energy increase.

B. 40 J per lift; 8 W and 20 W.

Review this choice. The work is correct, but the powers are reversed. Delivering the same energy in less time requires greater average power.

C. 40 J per lift; 20 W and 8 W.

Correct. Each hoist does mgh = (5)(10)(0.80) = 40 J of work. Dividing by 2.0 s and 5.0 s gives 20 W and 8 W, respectively.

D. 100 J per lift; 50 W and 20 W.

Review this choice. This overstates the energy change. Time changes the rate of delivery, not mgh for the specified mass and height.

The principle to carry forward

Equal endpoint energy changes can require different powers. Always distinguish one agent’s work from the net work of all forces.