Can an Object Have Kinetic Energy While Net Power Is Zero?

A car travels along a level road at a constant speed of 20 m/s. Its engine is still producing power, and air resistance is still removing energy. Yet the car’s kinetic energy isn’t changing.

How can the car have kinetic energy while its net power is zero?

Consider a 900 kg car. Its kinetic energy is:

\[ K=\frac12 mv^2 \]

\[ K=\frac12(900)(20)^2 =180\,000\text{ J} \]

The car has 180 kJ of kinetic energy. Now suppose the engine produces a forward force of 600 N, exactly balancing 600 N of resistance.

The engine supplies \(+12\,000\text{ W}\), while resistance removes \(12\,000\text{ W}\). The net power is zero.

Yes, an object can have kinetic energy while its net power is zero. The key is that kinetic energy measures an amount of energy, whereas power measures the rate at which that energy changes.

01Energy is an amount; power is a rate

Think about a water tank with water flowing into it and out of it at equal rates.

The tank can contain plenty of water even though its water level isn’t changing. The amount of water is not the same thing as the net rate of change.

Kinetic energy and power have a similar relationship:

• Kinetic energy tells you how much energy an object has because of its motion.
• Power tells you how quickly work is being done or energy is being transferred.
• Net power tells you the overall rate at which the object’s kinetic energy changes.

The distinction is important. An object doesn’t need to gain kinetic energy to receive energy from another source. It might receive energy at exactly the same rate that another force removes it.

For the car above, the engine supplies energy, but resistance removes it at the same rate. The car continues moving with unchanged kinetic energy.

The water-tank analogy only describes the relationship between an amount and its rate of change. Kinetic energy is not literally a fluid flowing through an object.

02Testing the idea: a trolley experiment

How could you demonstrate that an object can have positive kinetic energy and zero net power?

A useful experiment involves a trolley moving along a horizontal track while a motor pulls it forward against a controlled resistance.

Setting up the experiment

Imagine the following arrangement:

• A trolley of known mass sits on a horizontal track.
• A motor at the right-hand end pulls the trolley to the right through an inline force sensor.
• A controllable brake at the left-hand end resists the trolley’s motion. A second force sensor measures this opposing force.
• A motion sensor records the trolley’s velocity continuously.
• The trolley’s mass is measured using a balance.

Choose the direction to the right as positive.

The motor exerts a positive force. The brake exerts a negative force. The weight and normal reaction act vertically, perpendicular to the trolley’s horizontal motion.

What should the experiment show?

A first attempt might use only the force sensor connected to the motor.

This measures the motor’s power, but it doesn’t measure the net power on the trolley. The brake is also doing work, and ignoring it gives an incomplete result.

The revised arrangement includes the second force sensor. Now both the energy supplied and the energy removed can be measured.

The motion sensor also needs to record velocity over an interval. A single velocity reading cannot establish that the trolley is travelling at constant speed.

Suppose the experiment produces the following illustrative readings.

First, calculate the trolley’s kinetic energy:

\[ K=\frac12 mv^2 \]

\[ K=\frac12(0.80)(1.50)^2 \]

\[ \boxed{K=0.90\text{ J}} \]

The trolley has positive kinetic energy.

Now calculate the power supplied by the motor:

\[ P_{\text{motor}}=Fv \]

\[ P_{\text{motor}}=(0.60)(1.50) \]

\[ P_{\text{motor}}=+0.90\text{ W} \]

The brake removes energy at a rate of:

\[ P_{\text{brake}}=(-0.60)(1.50) \]

\[ P_{\text{brake}}=-0.90\text{ W} \]

Therefore:

\[ P_{\text{net}}=P_{\text{motor}}+P_{\text{brake}} \]

\[ P_{\text{net}}=0.90-0.90 \]

\[ \boxed{P_{\text{net}}=0\text{ W}} \]

The motor supplies energy at exactly the rate the brake removes it.

The trolley continues moving at 1.50 m/s, maintaining its 0.90 J of kinetic energy.

This is the important experimental distinction: zero net power does not mean that no energy is being transferred. Individual forces can transfer energy in opposite directions at equal rates.

The energy removed by the brake is ultimately transferred to thermal energy in the brake and its surroundings.

Assumptions and limitations: These readings represent an idealised model. The track is horizontal, the trolley is treated as a translating object, and all relevant horizontal resistance is included in the brake measurement. In a real experiment, the powers would only balance within the measurement uncertainty.

03The mathematical relationship between power and kinetic energy

The experiment gives us a useful physical picture. We can now express the relationship more precisely.

The instantaneous power delivered by a force is:

\[ P=Fv\cos\theta \]

where:

• \(P\) is power in watts (W),
• \(F\) is the force magnitude in newtons (N),
• \(v\) is the object’s speed in metres per second (m/s),
• \(\theta\) is the angle between the force and the velocity.

The angle matters because only the component of the force parallel to the velocity transfers energy into or out of the object’s kinetic energy.

This gives us three important cases:

When several forces act on an object, we add their individual powers, including their signs.

\[ P_{\text{net}}=\sum P_i \]

Equivalently:

\[ P_{\text{net}}=\vec F_{\text{net}}\cdot\vec v \]

This is the dot product of the net force and the velocity vectors.

For an object treated as a particle with constant mass, the work-energy theorem gives the instantaneous relationship:

\[ \boxed{P_{\text{net}}=\frac{dK}{dt}} \]

In words, net power is the instantaneous rate of change of kinetic energy.

This is the power form of the work-energy theorem.

It explains the original question directly:

\[ P_{\text{net}}=0 \]

means:

\[ \frac{dK}{dt}=0 \]

The kinetic energy isn’t changing at that instant.

It does not mean:

\[ K=0 \]

An object can have a substantial amount of kinetic energy while its rate of change is zero.

04Zero net force isn’t the only way to get zero net power

The trolley experiment involved balanced forces. The net force was zero, so the net power was also zero.

However, there’s another possibility that is particularly important in physics.

An object can experience a non-zero net force while its net power remains zero.

This happens when the net force is perpendicular to the object’s velocity.

Example: an object moving in a circle

Imagine a puck moving around a horizontal circular track at constant speed.

Its velocity is always tangent to the circular path. The centripetal force always points towards the centre of the circle.

These directions are perpendicular.

The net power is therefore:

\[ P_{\text{net}}=F_{\text{net}}v\cos90^\circ \]

\[ \boxed{P_{\text{net}}=0} \]

Yet the centripetal force is not zero.

It continuously changes the direction of the puck’s velocity without changing its speed.

Since kinetic energy depends on speed rather than direction,

\[ K=\frac12 mv^2 \]

the puck’s kinetic energy remains constant.

This is a useful distinction:

• Zero net force means zero acceleration.
• Zero net power means zero rate of change of kinetic energy.
• An object can have acceleration without changing its speed.

The second and third points are particularly important when studying circular motion.

05A particularly useful case: a projectile at its highest point

There is another situation where students often make the wrong prediction.

Imagine throwing a ball at an angle above the horizontal. At its highest point, its vertical velocity becomes zero.

Does that mean its total velocity is zero?

Not necessarily.

If the ball was thrown at an angle, it still has horizontal velocity at the highest point. Its speed is therefore still positive.

Assuming air resistance is negligible:

• The ball’s velocity at the highest point is horizontal.
• Gravity acts vertically downwards.
• The net force is perpendicular to the velocity.

Consequently, the net power is zero at that instant, even though the ball still has kinetic energy and a non-zero net force.

There is an important difference between this example and uniform circular motion.

For uniform circular motion, the net power is zero continuously because the centripetal force remains perpendicular to the velocity.

For a projectile, the net power is zero only at the highest point.

Immediately before reaching the highest point, gravity acts against the ball’s upward vertical motion, so its power is negative. Immediately afterwards, gravity acts in the same direction as the ball’s downward vertical motion, so its power is positive.

The ball’s kinetic energy decreases as it rises, reaches a minimum at the highest point, and increases as it falls.

06Instantaneous power versus average power

One final distinction can prevent a lot of mistakes in HSC exam questions.

When someone says that net power is zero, check whether they mean instantaneous power or average power over a time interval.

These statements are not interchangeable.

Average net power is calculated using:

\[ P_{\text{net,avg}}=\frac{\Delta K}{\Delta t} \]

For example, consider a projectile that returns to its original height.

If air resistance is negligible, its final speed equals its initial speed. Its initial and final kinetic energies are therefore equal.

The average net power over the entire flight is zero.

However, its kinetic energy decreases during the ascent and increases during the descent.

A zero average does not mean that the instantaneous net power was zero throughout the flight.

07How does efficiency fit into this?

There’s one more situation where students can confuse net power with another type of power: calculating the efficiency of a motor.

Suppose a motor lifts a load vertically at constant speed.

The load has positive kinetic energy because it is moving. Its kinetic energy remains constant because its speed doesn’t change.

The upward tension balances the downward gravitational force.

Therefore, the net power on the load is zero.

But the motor is still supplying energy. Where does that energy go?

It increases the gravitational potential energy of the load-Earth system.

Understanding the efficiency calculation

The standard efficiency convention is:

\[ \eta= \frac{P_{\text{useful output}}} {P_{\text{total input}}} \times100\% \]

Alternatively, you can use useful energy output divided by total energy input over the same interval.

The important decision is identifying what counts as the useful output and what counts as the input.

Consider a 5.0 kg load being lifted at a constant speed of 0.40 m/s. The motor receives 28 W of electrical power.

The useful output is the rate at which gravitational potential energy increases:

\[ P_{\text{useful}}=mgv \]

\[ P_{\text{useful}}=5.0(9.8)(0.40) \]

\[ P_{\text{useful}}=19.6\text{ W} \]

The motor’s efficiency is:

\[ \eta=\frac{19.6}{28}\times100\% \]

\[ \boxed{\eta=70\%} \]

The remaining 8.4 W is transferred to other energy stores, mainly thermal energy, in the motor and surrounding system.

Notice that the load itself has zero net power:

• The cable supplies \(+19.6\text{ W}\).
• Gravity contributes \(-19.6\text{ W}\).
• The net power on the load is zero.

Yet the motor’s useful output is 19.6 W.

Why? Because the useful output is the increase in gravitational potential energy, not the increase in kinetic energy.

Never assume that zero net power on an object means zero efficiency. Efficiency depends on the useful energy transfer and the total energy input, not simply the change in the object’s kinetic energy.

08A quick method for HSC Physics questions

When you encounter a question about kinetic energy and net power, use this sequence.

1. Identify the object and reference frame. Decide whose kinetic energy and power you are calculating. Use a consistent reference frame.
2. Identify the velocity. Power depends on the direction as well as the magnitude of velocity.
3. Consider each force separately. Determine whether it acts along, against, or perpendicular to the velocity.
4. Calculate signed powers. Use \(P=Fv\cos\theta\).
5. Add the powers. The sum gives the net power on the object.
6. Interpret the result. Use \(P_{\text{net}}=dK/dt\) to explain what happens to kinetic energy.

Keep these common misconceptions in mind:

For the connection between zero net force and constant velocity, see Newton’s First Law: Inertia and Zero Net Force Explained.

09Questions and solutions

The following questions progress from a straightforward power calculation to problems involving circular motion, efficiency, and energy transfer between interacting objects.

Question 1

A 1.5 kg trolley moves horizontally at a constant speed of 2.0 m/s. A motor exerts a forward force of 3.0 N, while friction exerts a backward force of 3.0 N.

Treat the trolley as a translating object.

Calculate:

(a) The kinetic energy of the trolley.

(b) The power supplied by the motor.

(c) The power removed by friction.

(d) The net power on the trolley.

Explain how the trolley can have kinetic energy while its net power is zero.

Solution 1

The trolley has 3.0 J of kinetic energy, the motor supplies 6.0 W, friction removes 6.0 W, and the net power is zero.

(a) Kinetic energy

Use:

\[ K=\frac12 mv^2 \]

Substitute:

\[ K=\frac12(1.5)(2.0)^2 \]

\[ \boxed{K=3.0\text{ J}} \]

The trolley has positive kinetic energy because it is moving.

(b) Power supplied by the motor

The motor’s force acts in the direction of motion.

\[ P_{\text{motor}}=Fv \]

\[ P_{\text{motor}}=(3.0)(2.0) \]

\[ \boxed{P_{\text{motor}}=+6.0\text{ W}} \]

(c) Power removed by friction

Friction acts opposite to the direction of motion.

\[ P_{\text{friction}}=-Fv \]

\[ P_{\text{friction}}=-(3.0)(2.0) \]

\[ \boxed{P_{\text{friction}}=-6.0\text{ W}} \]

The negative sign indicates that friction removes kinetic energy from the trolley.

(d) Net power

Add the individual powers:

\[ P_{\text{net}}=P_{\text{motor}}+P_{\text{friction}} \]

\[ P_{\text{net}}=6.0-6.0 \]

\[ \boxed{P_{\text{net}}=0\text{ W}} \]

The motor supplies energy at exactly the rate friction removes it.

The trolley therefore maintains its constant speed and its 3.0 J of kinetic energy.

The tempting mistake is to interpret zero net power as zero kinetic energy. The actual condition is that the kinetic energy isn’t changing.

Question 2

A 0.20 kg ball is projected at 25 m/s at an angle of \(36.9^\circ\) above the horizontal.

Assume air resistance is negligible and take \(g=9.8\text{ m/s}^2\).

At the highest point of its trajectory, calculate:

(a) Its speed.

(b) Its kinetic energy.

(c) The magnitude and direction of the net force.

(d) Its instantaneous net power.

Explain why the net power is zero even though the net force is non-zero.

Solution 2

At the highest point, the ball has a speed of 20 m/s, kinetic energy of 40 J, a net force of 1.96 N downwards, and zero instantaneous net power.

(a) Speed at the highest point

Resolve the initial velocity horizontally:

\[ v_x=v_0\cos\theta \]

\[ v_x=25\cos36.9^\circ \]

\[ v_x=20\text{ m/s} \]

At the highest point, the vertical velocity is zero.

The horizontal velocity remains unchanged, so:

\[ \boxed{v=20\text{ m/s}} \]

(b) Kinetic energy

\[ K=\frac12 mv^2 \]

\[ K=\frac12(0.20)(20)^2 \]

\[ \boxed{K=40\text{ J}} \]

The ball still has substantial kinetic energy at the apex.

(c) Net force

Gravity is the only force acting on the ball.

\[ F_{\text{net}}=mg \]

\[ F_{\text{net}}=0.20(9.8) \]

\[ \boxed{F_{\text{net}}=1.96\text{ N downward}} \]

The net force is clearly not zero.

(d) Instantaneous net power

At the apex, the velocity is horizontal and gravity is vertical.

The angle between them is \(90^\circ\).

\[ P_{\text{net}}=F_{\text{net}}v\cos90^\circ \]

\[ \boxed{P_{\text{net}}=0\text{ W}} \]

The important point is that a non-zero net force does not necessarily produce a change in speed.

Here, gravity changes the vertical component of the velocity. At the exact highest point, that component is zero, so gravity has no component along the instantaneous velocity.

Immediately before the apex, gravity removes kinetic energy. Immediately afterwards, it increases kinetic energy.

The zero-power condition occurs at one instant only.

Question 3

A 1200 kg car travels around a flat circular track of radius 80 m at an instantaneous speed of 20 m/s.

At a particular instant:

• The engine exerts a forward tangential force of 2500 N.
• Air resistance and other tangential resistance total 700 N backwards.
• The tyres provide the inward force required for circular motion.

Treat the car as a point mass and ignore rotational kinetic energy.

Calculate:

(a) The net tangential force.

(b) The centripetal force.

(c) The magnitude of the resultant net force.

(d) The net power on the car.

Explain why multiplying the magnitude of the resultant net force by the speed gives an incorrect answer for net power.

Solution 3

The net tangential force is 1800 N, the centripetal force is 6000 N, and the resultant net force has a magnitude of approximately 6264 N. The net power is \(+36\,000\text{ W}\), or 36 kW.

(a) Net tangential force

The engine acts forwards, while resistance acts backwards.

\[ F_t=2500-700 \]

\[ \boxed{F_t=1800\text{ N}} \]

This force acts along the direction of motion and increases the car’s speed.

(b) Centripetal force

The car also needs an inward force to follow its circular path.

\[ F_c=\frac{mv^2}{r} \]

Substitute:

\[ F_c=\frac{1200(20)^2}{80} \]

\[ \boxed{F_c=6000\text{ N}} \]

This force acts towards the centre of the circle.

It is perpendicular to the car’s instantaneous velocity.

(c) Resultant net force

The tangential and centripetal forces are perpendicular.

Therefore, use Pythagoras’ theorem:

\[ F_{\text{net}}= \sqrt{F_t^2+F_c^2} \]

\[ F_{\text{net}}= \sqrt{1800^2+6000^2} \]

\[ \boxed{F_{\text{net}}\approx6264\text{ N}} \]

The resultant force is not parallel to the car’s velocity.

(d) Net power

The tangential force contributes to the change in speed.

The centripetal force is perpendicular to the velocity and contributes zero power.

Therefore:

\[ P_{\text{net}}=F_tv \]

\[ P_{\text{net}}=1800(20) \]

\[ \boxed{P_{\text{net}}=36\,000\text{ W}} \]

We can also check this using the individual powers:

\[ P_{\text{engine}}=2500(20)=50\,000\text{ W} \]

\[ P_{\text{resistance}}=-700(20)=-14\,000\text{ W} \]

The inward centripetal force contributes zero power.

Hence:

\[ P_{\text{net}}=50\,000-14\,000 \]

\[ \boxed{P_{\text{net}}=36\text{ kW}} \]

A common mistake is to calculate:

\[ P=F_{\text{net}}v \]

using the magnitude of the resultant force.

That would give approximately 125 kW, which is incorrect.

The formula \(P=Fv\) without an angle assumes that the force is parallel to the velocity. Here, only the tangential component contributes to net power.

This is why the dot-product form is essential:

\[ P_{\text{net}}=\vec F_{\text{net}}\cdot\vec v \]

Question 4

An 80 kg load is lifted vertically by an electric motor at a constant speed of 0.50 m/s.

The motor receives 500 W of electrical power.

Assume \(g=9.8\text{ m/s}^2\), and ignore changes in rotational kinetic energy.

Calculate:

(a) The kinetic energy of the load.

(b) The net power acting on the load.

(c) The useful mechanical output power of the lifting system.

(d) The efficiency of the motor.

(e) The rate at which energy is transferred to other stores rather than the useful gravitational potential energy output.

Explain why the useful output power is not the same as the net power on the load.

Solution 4

The load has 10 J of kinetic energy, its net power is zero, the useful output power is 392 W, and the motor efficiency is 78.4%. The remaining 108 W is transferred to other energy stores.

(a) Kinetic energy

\[ K=\frac12 mv^2 \]

\[ K=\frac12(80)(0.50)^2 \]

\[ \boxed{K=10\text{ J}} \]

The load has positive kinetic energy.

(b) Net power on the load

The load travels upwards at constant speed.

Therefore, its acceleration is zero, and the upward tension balances its weight.

\[ T=mg \]

\[ T=80(9.8)=784\text{ N} \]

The power supplied by the cable is:

\[ P_T=Tv \]

\[ P_T=784(0.50)=392\text{ W} \]

Gravity acts opposite to the direction of motion:

\[ P_g=-mgv \]

\[ P_g=-392\text{ W} \]

Therefore:

\[ P_{\text{net}}=392-392 \]

\[ \boxed{P_{\text{net}}=0\text{ W}} \]

The load’s kinetic energy remains constant.

(c) Useful mechanical output power

The useful output is the rate at which gravitational potential energy increases.

\[ P_{\text{useful}}=mgv \]

\[ P_{\text{useful}}=80(9.8)(0.50) \]

\[ \boxed{P_{\text{useful}}=392\text{ W}} \]

(d) Efficiency

The total electrical input is 500 W.

\[ \eta= \frac{P_{\text{useful}}}{P_{\text{input}}} \times100\% \]

\[ \eta=\frac{392}{500}\times100\% \]

\[ \boxed{\eta=78.4\%} \]

(e) Rate of energy transfer to other stores

The input power is 500 W, while the useful output is 392 W.

\[ P_{\text{other}}=500-392 \]

\[ \boxed{P_{\text{other}}=108\text{ W}} \]

The tempting mistake is to use the load’s net power as the useful output.

That would give an efficiency of zero, which is clearly inconsistent with the fact that the motor is successfully raising the load.

The distinction is the energy store being considered.

The load’s kinetic energy stays constant, but the gravitational potential energy of the load-Earth system increases. Efficiency measures the useful output relative to the total input, not simply the rate of change of kinetic energy.

Question 5

Two carts, A and B, move along a horizontal, frictionless track. They are connected by a light, ideal spring bumper.

Initially, the spring is uncompressed.

The following information is provided:

• Cart A has a mass of 1.0 kg and moves to the right at 4.0 m/s.
• Cart B has a mass of 3.0 kg and is initially stationary.
• The spring constant is 300 N/m.
• There are no external horizontal forces, and no energy is dissipated.

As cart A catches cart B, the spring compresses. At maximum compression, the two carts momentarily have the same velocity.

Calculate:

(a) The common velocity of the carts at maximum compression.

(b) The maximum compression of the spring.

(c) The kinetic energy of each cart at maximum compression.

(d) The power delivered by the spring to each cart at that instant.

(e) The combined instantaneous power delivered by the spring to both carts.

A student makes the following argument:

“The spring forces on the two carts are equal and opposite, according to Newton’s third law. Therefore, the spring’s combined power on the carts must be zero throughout the collision, meaning the total kinetic energy of the carts never changes.”

Evaluate this argument. Explain precisely when the spring’s combined power is zero, why the student’s reasoning fails during compression, and how energy is transferred between the carts and the spring.

Solution 5

At maximum compression, both carts move right at 1.0 m/s. The spring is compressed by 0.20 m, and the carts have kinetic energies of 0.50 J and 1.50 J respectively.

The spring delivers -60 W to cart A and +60 W to cart B. Its combined instantaneous power on the two carts is zero at maximum compression, but it is not zero throughout the collision.

The student’s argument is incorrect because equal and opposite forces do not necessarily produce equal and opposite powers.

(a) Common velocity at maximum compression

At maximum compression, the relative velocity of the carts is zero.

They must therefore be moving together momentarily at the same velocity.

Since there are no external horizontal forces, horizontal momentum is conserved.

Initial momentum:

\[ p_i=m_Au_A+m_Bu_B \]

\[ p_i=(1.0)(4.0)+(3.0)(0) \]

\[ p_i=4.0\text{ kg m/s} \]

Let the common velocity at maximum compression be \(v\).

Conservation of momentum gives:

\[ 4.0=(1.0+3.0)v \]

\[ \boxed{v=1.0\text{ m/s to the right}} \]

(b) Maximum compression of the spring

First, calculate the initial kinetic energy of the carts.

\[ K_i=\frac12m_Au_A^2+\frac12m_Bu_B^2 \]

\[ K_i=\frac12(1.0)(4.0)^2 \]

\[ K_i=8.0\text{ J} \]

At maximum compression, both carts move at 1.0 m/s.

Their combined kinetic energy is:

\[ K_f=\frac12(m_A+m_B)v^2 \]

\[ K_f=\frac12(4.0)(1.0)^2 \]

\[ K_f=2.0\text{ J} \]

Since mechanical energy is conserved, the decrease in kinetic energy becomes elastic potential energy in the spring.

\[ E_{\text{spring}}=8.0-2.0 \]

\[ E_{\text{spring}}=6.0\text{ J} \]

For an ideal spring:

\[ E_{\text{spring}}=\frac12kx^2 \]

Substitute:

\[ 6.0=\frac12(300)x^2 \]

\[ x^2=\frac{12}{300} \]

\[ x^2=0.040 \]

\[ \boxed{x=0.20\text{ m}} \]

The maximum compression is 0.20 m.

(c) Kinetic energy of each cart

For cart A:

\[ K_A=\frac12m_Av^2 \]

\[ K_A=\frac12(1.0)(1.0)^2 \]

\[ \boxed{K_A=0.50\text{ J}} \]

For cart B:

\[ K_B=\frac12m_Bv^2 \]

\[ K_B=\frac12(3.0)(1.0)^2 \]

\[ \boxed{K_B=1.50\text{ J}} \]

Both carts have positive kinetic energy at maximum compression.

(d) Power delivered by the spring to each cart

First, calculate the spring force at maximum compression.

Hooke’s law gives:

\[ F=kx \]

\[ F=300(0.20) \]

\[ F=60\text{ N} \]

The spring pushes the carts in opposite directions.

It pushes cart A to the left and cart B to the right.

For cart A, the force is opposite to its velocity:

\[ P_A=-Fv \]

\[ P_A=-60(1.0) \]

\[ \boxed{P_A=-60\text{ W}} \]

For cart B, the force is in the same direction as its velocity:

\[ P_B=Fv \]

\[ P_B=60(1.0) \]

\[ \boxed{P_B=+60\text{ W}} \]

At this instant, the spring removes kinetic energy from cart A and transfers it to cart B at equal rates.

(e) Combined instantaneous power

Add the powers delivered to both carts:

\[ P_{\text{spring,total}}=P_A+P_B \]

\[ P_{\text{spring,total}}=-60+60 \]

\[ \boxed{P_{\text{spring,total}}=0\text{ W}} \]

The spring’s combined power is zero at maximum compression.

Now consider the student’s argument.

Newton’s third law tells us that the spring forces on the two carts are equal in magnitude and opposite in direction.

However, power depends on both force and the velocity of the object experiencing that force.

In general, the spring’s combined power on the two carts is:

\[ P_{\text{total}}=-Fv_A+Fv_B \]

Rearranging:

\[ P_{\text{total}}=F(v_B-v_A) \]

During compression, cart A is moving faster than cart B.

Therefore:

\[ v_A>v_B \]

This means:

\[ P_{\text{total}}<0 \]

The spring removes kinetic energy from the combined carts and stores it as elastic potential energy.

At maximum compression:

\[ v_A=v_B \]

Therefore:

\[ P_{\text{total}}=0 \]

This is the instant when the spring’s combined power on the carts becomes zero.

After maximum compression, cart B moves faster than cart A as the spring expands.

Now:

\[ v_B>v_A \]

Therefore:

\[ P_{\text{total}}>0 \]

The spring releases its stored elastic potential energy, increasing the combined kinetic energy of the carts.

The student’s mistake was assuming that equal and opposite forces must always have equal and opposite powers.

That is only true in this arrangement when the carts have equal velocities. During most of the compression and expansion, their velocities differ.

This example reveals an important distinction: Newton’s third law concerns forces, while power depends on the force acting on each object’s own velocity.

The combined power can be zero even when the individual powers are non-zero. It can also be non-zero even when the combined force is zero.

The most useful next step is to connect instantaneous power to a power-time graph. The signed area under a net-power-versus-time graph gives the change in kinetic energy:

\[ \Delta K=\int P_{\text{net}}\,dt \]

That relationship brings together the two ideas in this article: how much kinetic energy an object has, and how quickly that energy changes.