A spring is stretched slowly from its natural length to an extension x0x_0. What is the work done by the spring itself over that stretch?

Two graphs, one stretch

The hand and the spring pull on each other with equal and opposite forces throughout. Their force-extension graphs are therefore mirror images. The two areas are equal in size and opposite in sign.

Same stretch, same size of area, opposite sides of the axis. The hand puts tfrac12kx2 in and the spring takes exactly that much back out of the motion, which is why nothing speeds up.
Same stretch, same size of area, opposite sides of the axis. The hand puts 12kx2\tfrac{1}{2}kx^2 in and the spring takes exactly that much back out of the motion, which is why nothing speeds up.

That is why a slow stretch leaves the block with no kinetic energy. The two works cancel exactly. What survives is stored in the spring, and the next chapter gives that store a name.

A spring is stretched from an extension of 2 cm to an extension of 4 cm. How does the work needed compare with the work needed for the first 2 cm?

03

Work Done by Common Forces

Introduction

A few forces appear again and again in mechanics. This section works out the work done by each of them. These results are the building blocks for solving more complex energy problems.

Work Done by Gravitational Force

The gravitational force, F⃗g=mg⃗\vec{F}_g = m\vec{g}, is constant in magnitude and direction near the Earth's surface. The work done by gravity depends only on the vertical displacement of the object, not on the path it takes. Let the vertical displacement be Δy=yf−yi\Delta y = y_f - y_i, where 'up' is the positive direction.

The work done by gravity is:

Wg=−mgΔy=−mg(yf−yi)W_g = -mg\Delta y = -mg(y_f - y_i)

  • If an object moves downwards (yf<yiy_f < y_i), Δy\Delta y is negative. The work done by gravity is positive.
  • If an object moves upwards (yf>yiy_f > y_i), Δy\Delta y is positive, and the work done by gravity is negative.
  • If an object moves horizontally, the vertical displacement is zero, and the work done by gravity is zero.

Gravity's work depends only on the change in height, and the two trips cancel. This is the first hint that gravity is a conservative force.
Gravity's work depends only on the change in height, and the two trips cancel. This is the first hint that gravity is a conservative force.

A 2 kg book is lifted 1.5 m upwards. What is the work done by the gravitational force? (Use g = 9.8 m/s²)

Work Done by Normal Force

The normal force (N⃗\vec{N}) is a contact force acting perpendicular to the surface of contact. Often, the normal force does zero work.

  • Zero Work: If an object slides along a fixed surface (horizontal or inclined), the normal force is always perpendicular (θ=90∘\theta = 90^\circ) to the displacement. Therefore, WN=0W_N = 0. This is the most common scenario.
  • Non-Zero Work: If the surface itself is moving, the normal force can do work. For example, consider an elevator accelerating upwards. The floor exerts a normal force on you and moves in the same direction, doing positive work.

For a system of two interacting bodies moving together, the net work done by the internal normal forces (action-reaction pair) is always zero.

"The normal force never does work" is a habit, not a law. It does no work only while the surface itself does not move along its own normal.
"The normal force never does work" is a habit, not a law. It does no work only while the surface itself does not move along its own normal.

You stand on the floor of a lift that is accelerating upwards. Does the normal force from the floor do work on you?

Work Done by Friction

Friction is a dissipative force that opposes motion or attempted motion. The work done by friction is almost always negative, as it removes mechanical energy from a system.

Kinetic Friction

The force of kinetic friction, fk=μkNf_k = \mu_k N, acts opposite to the direction of motion (θ=180∘\theta = 180^\circ). If an object slides a distance dd, the work done by kinetic friction is:

Wf=−fkd=−μkNdW_f = -f_k d = -\mu_k N d

Kinetic friction opposes the relative sliding by definition, so the angle is always 180circ and the work is always negative.
Kinetic friction opposes the relative sliding by definition, so the angle is always 180∘180^\circ and the work is always negative.

Static Friction

The force of static friction (fsf_s) prevents motion. If the point of application of the force does not move, the work done by static friction is zero. For example, for a block at rest on an incline, static friction does no work. However, static friction can do work. When you walk, the static friction from the ground on your shoes pushes you forward, doing positive work. When a car accelerates, static friction from the road on the tires does positive work on the car. This increases its kinetic energy.

Static friction is what accelerates the crate, and the crate moves the way it points, so here friction is the force feeding energy in.
Static friction is what accelerates the crate, and the crate moves the way it points, so here friction is the force feeding energy in.

A box is pulled 10 m across a rough horizontal floor (μk=0.3\mu_k = 0.3). The box has a mass of 20 kg. What is the work done by kinetic friction? (Use g = 9.8 m/s²)

A crate sits on the flat bed of a truck without slipping. The truck accelerates forwards. What work does the friction on the crate do?

Work Done by Tension

Tension (TT) is a force transmitted through a string, rope, or cable. The work done by tension depends on the angle between the tension force and the displacement.

  • If a block is pulled by a horizontal rope, tension does positive work.
  • In an Atwood machine, tension does positive work on the rising mass and negative work on the falling mass.

Consider a system connected by an inextensible string. The net work done by the internal tension forces can be non-zero if the parts of the system move differently. For example, take an accelerating Atwood machine. The tension force is the same on both masses. They move the same distance in opposite directions, so the net work done by tension on the system is zero. However, the rope may pass over a massive, rotating pulley. Then the work done by tension on the pulley is non-zero, and the pulley gains rotational kinetic energy.

The string is inextensible, so the two ends move equal distances; the tension is the same at both, so its two contributions cancel exactly.
The string is inextensible, so the two ends move equal distances; the tension is the same at both, so its two contributions cancel exactly.

Three of the four have a condition; kinetic friction has none, which is exactly what marks it out as the non-conservative one.
Three of the four have a condition; kinetic friction has none, which is exactly what marks it out as the non-conservative one.

Worked Example

Illustrative Example: Net Work on a Pulled Block

A 5 kg block is pulled up a frictionless ramp inclined at 30° by a constant force of 40 N parallel to the ramp. The block moves a distance of 3 m along the ramp. Calculate the work done by each force and the net work done on the block.

Three forces act. N is perpendicular to the 3 m displacement and does nothing. F is along the displacement. Gravity gets in through the 1.5,m of height the block gains.
Three forces act. NN is perpendicular to the 3 m displacement and does nothing. FF is along the displacement. Gravity gets in through the 1.5 m1.5\,\mathrm{m} of height the block gains.

Solution:

The forces acting on the block are: the applied force (FappF_{app}), gravity (FgF_g). The normal force (NN).

  1. Work done by Applied Force (WappW_{app}):
    The force is parallel to the displacement, so θ=0∘\theta = 0^\circ.
    Wapp=Fappdcos⁡(0∘)=(40 N)(3 m)(1)=120 JW_{app} = F_{app} d \cos(0^\circ) = (40 \text{ N})(3 \text{ m})(1) = 120 \text{ J}
  2. Work done by Normal Force (WNW_N):
    The normal force is perpendicular to the ramp, so it is perpendicular to the displacement. θ=90∘\theta = 90^\circ.
    WN=0 JW_N = 0 \text{ J}
  3. Work done by Gravity (WgW_g):
    Gravity acts vertically downwards. The vertical displacement is Δy=dsin⁡(30∘)=(3 m)(0.5)=1.5 m\Delta y = d \sin(30^\circ) = (3 \text{ m})(0.5) = 1.5 \text{ m} upwards.
    Wg=−mgΔy=−(5 kg)(9.8 m/s2)(1.5 m)=−73.5 JW_g = -mg\Delta y = -(5 \text{ kg})(9.8 \text{ m/s}^2)(1.5 \text{ m}) = -73.5 \text{ J}
  4. Net Work Done (WnetW_{net}):
    The net work is the sum of the work done by all forces.
    Wnet=Wapp+WN+Wg=120 J+0 J+(−73.5 J)=46.5 JW_{net} = W_{app} + W_N + W_g = 120 \text{ J} + 0 \text{ J} + (-73.5 \text{ J}) = 46.5 \text{ J}

The net work done on the block is 46.5 J. By the Work-Energy Theorem, this positive net work increases the block's kinetic energy.

04

Work Done by Internal Forces

Understanding Internal Forces and Work

Internal forces always occur in equal and opposite action-reaction pairs, by Newton's Third Law. For a system of particles, the vector sum of all internal forces is therefore always zero (∑F⃗int=0\sum \vec{F}_{int} = 0). It is tempting, but incorrect, to assume that the total work done by these internal forces must also be zero.

Work is defined by the dot product F⃗⋅ds⃗\vec{F} \cdot d\vec{s}. The forces in an action-reaction pair are equal and opposite. The displacements of their points of application, however, are not necessarily the same. If the points where the two forces act undergo different displacements, the work done by them will not cancel out.

Here the two points of application share one displacement, so +FABd and -FABd really do sum to zero. The explosion differs in the displacements, not in the forces.
Here the two points of application share one displacement, so +FABd+F_{AB}d and −FABd-F_{AB}d really do sum to zero. The explosion differs in the displacements, not in the forces.

Worked Example

Illustrative Example: An Exploding Bomb

Consider a bomb of mass MM initially at rest. It explodes into two fragments, m1m_1 and m2m_2. The forces of the explosion are internal to the bomb system. The force on fragment 1 by fragment 2 (F⃗12\vec{F}_{12}) is equal and opposite to the force on 2 by 1 (F⃗21\vec{F}_{21}).

vec F12 = -vec F21, but the two points of application move in opposite directions too, so the works add instead of cancelling.
F⃗12=−F⃗21\vec F_{12} = -\vec F_{21}, but the two points of application move in opposite directions too, so the works add instead of cancelling.

Although the forces are equal and opposite, the fragments move apart. Both forces therefore do positive work on their respective fragments. For instance, F⃗21\vec{F}_{21} pushes m2m_2 to the right, and m2m_2 displaces to the right, so W21>0W_{21} > 0. Similarly, F⃗12\vec{F}_{12} pushes m1m_1 to the left, and m1m_1 displaces to the left, so W12>0W_{12} > 0.

The total work done by internal forces is Wint=W12+W21>0W_{int} = W_{12} + W_{21} > 0. This positive net work done by internal forces is responsible for the increase in the system's kinetic energy. The initial kinetic energy was zero, and the final kinetic energy is large and positive. This energy came from the chemical potential energy stored in the bomb's material. The work done by the explosive forces converted that store into the kinetic energy of the fragments. This shows that internal forces can do net work and change the kinetic energy of a system.

Where the kinetic energy came from

Positive internal work is not energy appearing from nowhere. The explosive carried a chemical store. The work done by the internal forces emptied that store into the motion of the fragments.

Net internal work is not energy from nowhere: it is a store inside the system being spent. The total energy is unchanged; only its form is.
Net internal work is not energy from nowhere: it is a store inside the system being spent. The total energy is unchanged; only its form is.

The total energy of the system never changed. What changed is which account it sits in. This chapter tracks only the mechanical accounts, kinetic and potential.

The same reasoning settles a question that looks paradoxical. When you jump, the floor is the only external force that can push you up. Yet it does no work on you at all.

The floor supplies the external force that changes the momentum and does no work at all, because its point of application never moves. The kinetic energy comes from an internal chemical store.
The floor supplies the external force that changes the momentum and does no work at all, because its point of application never moves. The kinetic energy comes from an internal chemical store.

The floor's push acts at your feet. While your feet are on the floor they do not move. The force is external and the work is internal. That is why the two questions have to be asked separately.

In the exploding bomb example, is the net work done by the internal forces on the system positive, negative, or zero?

Two identical blocks are pushed along a smooth floor together, in contact, by a single force on the back one. What is the total work done by the contact forces between the two blocks?

05

Kinetic Energy and the Work-Energy Theorem

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. So far we have discussed work as a measure of energy transfer. Kinetic energy is the form of energy that transfer most often changes. The term "kinetic" comes from the Greek word kinetos, meaning "motion". For a non-rotating, rigid object of mass mm moving with speed vv, the kinetic energy (KK) is defined as:

K=12mv2\qquad K = \frac{1}{2}mv^2

Key points about kinetic energy:

  • Scalar Quantity: Like work, kinetic energy is a scalar quantity. It has magnitude but no direction.
  • Depends on Speed, Not Velocity: Since v2v^2 is always non-negative, kinetic energy is always positive or zero. An object has kinetic energy regardless of the direction of its motion.
  • Units: The SI unit of kinetic energy is the Joule (J), the same as work. Checking the units: kg⋅(m/s)2=(kg⋅m/s2)⋅m=N⋅m=J\text{kg} \cdot (\text{m/s})^2 = (\text{kg} \cdot \text{m/s}^2) \cdot \text{m} = \text{N} \cdot \text{m} = \text{J}.
  • Proportional to Mass and Speed Squared: Kinetic energy is directly proportional to the mass of the object and, more significantly, to the square of its speed. This means doubling the speed of an object quadruples its kinetic energy.

This is why stopping distance rises so steeply with speed: at twice the speed there is four times as much kinetic energy to remove.
This is why stopping distance rises so steeply with speed: at twice the speed there is four times as much kinetic energy to remove.

Squaring the speed destroys the sign, so kinetic energy is a positive scalar. Two bodies heading straight at each other both carry positive K.
Squaring the speed destroys the sign, so kinetic energy is a positive scalar. Two bodies heading straight at each other both carry positive KK.

If a car's speed doubles, by what factor does its kinetic energy increase?

The Work-Energy Theorem

The relationship between work and kinetic energy is one of the most important principles in mechanics. The Work-Energy Theorem states that the net work done on an object by all forces is equal to the change in its kinetic energy.

The net work (WnetW_{net}) is the algebraic sum of the work done by all individual forces acting on the object. Examples are applied forces, gravity, friction. The normal force.

Wnet=ΔK=Kf−Ki\qquad W_{net} = \Delta K = K_f - K_i

Where:

  • WnetW_{net} is the total work done on the object.
  • ΔK\Delta K is the change in kinetic energy.
  • Kf=12mvf2K_f = \frac{1}{2}mv_f^2 is the final kinetic energy (with final speed vfv_f).
  • Ki=12mvi2K_i = \frac{1}{2}mv_i^2 is the initial kinetic energy (with initial speed viv_i).

The theorem tells us:

  • If the net work done on an object is positive (Wnet>0W_{net} > 0), its kinetic energy increases (Kf>KiK_f > K_i), and it speeds up.
  • If the net work done on an object is negative (Wnet<0W_{net} < 0), its kinetic energy decreases (Kf<KiK_f < K_i), and it slows down.
  • If the net work done on an object is zero (Wnet=0W_{net} = 0), its kinetic energy remains constant (Kf=KiK_f = K_i), and its speed does not change.

Everything the derivation needs is here: a constant net force, a straight displacement, and the two speeds at the ends of it.
Everything the derivation needs is here: a constant net force, a straight displacement, and the two speeds at the ends of it.

Wnet = Kf - Ki is a statement about one bar changing height. Nothing in it cares which forces did the work.
Wnet=Kf−KiW_{\text{net}} = K_f - K_i is a statement about one bar changing height. Nothing in it cares which forces did the work.

Derivation of the Work-Energy Theorem (for a constant net force in 1D)

Consider a particle of mass mm moving in a straight line (the x-axis) acted on by a constant net force FnetF_{net} directed along the line of motion. The particle accelerates from an initial speed viv_i to a final speed vfv_f over a displacement dd.

1. From Newton's Second Law, the net force is Fnet=maF_{net} = ma, where aa is the constant acceleration.

2. The work done by this net force is Wnet=Fnet⋅d=(ma)dW_{net} = F_{net} \cdot d = (ma)d.

3. From kinematics, we have the equation relating initial and final velocities, acceleration, and displacement: vf2=vi2+2adv_f^2 = v_i^2 + 2ad.

4. We can rearrange this equation to solve for the product adad: ad=vf2−vi22ad = \frac{v_f^2 - v_i^2}{2}.

5. Now, substitute this expression for adad back into the equation for work:

Wnet=m(ad)=m(vf2−vi22)\qquad W_{net} = m(ad) = m \left( \frac{v_f^2 - v_i^2}{2} \right)

Wnet=12mvf2−12mvi2\qquad W_{net} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Wnet=Kf−Ki=ΔK\qquad W_{net} = K_f - K_i = \Delta K

This proves the theorem for a constant net force. A more general proof using integration can be used for variable forces.

The theorem needs the net work

The theorem is about the net work. The commonest way to misapply it is to use the work of whichever force the question mentions. Add the works of all of them. Include the ones that contribute zero, so that you have checked.

Add the works, not the forces. Two of these four contribute nothing because they are perpendicular to the displacement.
Add the works, not the forces. Two of these four contribute nothing because they are perpendicular to the displacement.

Two of the four forces here are perpendicular to the displacement and contribute nothing. That is worth writing down rather than assuming.

If the net work done on an object is negative, what happens to its speed?

A block is dragged at a perfectly constant speed across a rough floor by a horizontal rope. What is the net work done on the block?

Worked Example

Illustrative Example: Work to Stop a Car

A car of mass 1000 kg is traveling at a speed of 20 m/s. The driver applies the brakes, and the car skids to a stop. Calculate the work done by the friction force to bring the car to rest.

The brakes must remove all of tfrac12mv2, so Fd = tfrac12mv2 fixes the distance without ever finding the acceleration.
The brakes must remove all of 12mv2\tfrac{1}{2}mv^2, so Fd=12mv2Fd = \tfrac{1}{2}mv^2 fixes the distance without ever finding the acceleration.

Solution:

We can use the Work-Energy Theorem. The only force doing work during the skid is the kinetic friction force from the road. Therefore, the net work is the work done by friction, Wnet=WfrictionW_{net} = W_{friction}.

Given:

  • Mass, m=1000m = 1000 kg
  • Initial speed, vi=20v_i = 20 m/s
  • Final speed, vf=0v_f = 0 m/s (since the car comes to a stop)

1. Calculate the initial kinetic energy:

Ki=12mvi2=12(1000 kg)(20 m/s)2=12(1000)(400)=200,000 J\qquad K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(1000 \, \text{kg})(20 \, \text{m/s})^2 = \frac{1}{2}(1000)(400) = 200,000 \, \text{J}

2. Calculate the final kinetic energy:

Kf=12mvf2=12(1000)(0)2=0 J\qquad K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(1000)(0)^2 = 0 \, \text{J}

3. Apply the Work-Energy Theorem:

Wnet=ΔK=Kf−Ki\qquad W_{net} = \Delta K = K_f - K_i

Wfriction=0 J−200,000 J=−200,000 J\qquad W_{friction} = 0 \, \text{J} - 200,000 \, \text{J} = -200,000 \, \text{J} or -200 kJ.

The work done by friction is -200 kJ. The negative sign is expected. Friction is a non-conservative force that removes kinetic energy from the system (dissipating it as heat) to slow the car down.

The same square law, read as a distance. It is the work-energy theorem rearranged, and it is why a small speed increase is not a small risk.
The same square law, read as a distance. It is the work-energy theorem rearranged, and it is why a small speed increase is not a small risk.

06

Power

Introduction to Power

Power (PP) is the rate at which work is done or, equivalently, the rate at which energy is transferred or transformed. In physics it often matters not only how much work is done but also how quickly it is done. Power measures this rate.

Average Power

If an amount of work WW is done over a time interval Δt\Delta t, the average power (PavgP_{avg}) is given by:

Pavg=WΔt\qquad P_{avg} = \frac{W}{\Delta t}

Since work represents an energy transfer, power is also the rate of energy transfer. If an amount of energy ΔE\Delta E is transferred in time Δt\Delta t, the average power is:

Pavg=ΔEΔt\qquad P_{avg} = \frac{\Delta E}{\Delta t}

Work asks how much energy moved; power asks how fast. The two loads end up identically higher, and only one of the lifters is out of breath.
Work asks how much energy moved; power asks how fast. The two loads end up identically higher, and only one of the lifters is out of breath.

Instantaneous Power

To find the power at a specific moment in time, we consider an infinitesimally small time interval dtdt. The instantaneous power (PP) is the limit of the average power as the time interval approaches zero. It is the time derivative of work:

P=dWdt\qquad P = \frac{dW}{dt}

The SI unit of power is the Watt (W), named after the Scottish engineer James Watt. One watt is equal to one joule per second:

1 Watt=1 Joule/second(1 W=1 J/s)\qquad 1 \, \text{Watt} = 1 \, \text{Joule/second} \quad (1 \, \text{W} = 1 \, \text{J/s})

Another common unit of power, especially in the context of engines, is horsepower (hp). The conversion is approximately 1 hp≈746 W1 \, \text{hp} \approx 746 \, \text{W}.

Average power is the slope between two points on this curve; instantaneous power is the slope at one point. Same graph, two questions.
Average power is the slope between two points on this curve; instantaneous power is the slope at one point. Same graph, two questions.

If two people lift identical weights to the same height, but the first person does it in half the time, which person exerts more power?

Power in Terms of Force and Velocity

We can derive a useful expression for instantaneous power in terms of force and velocity. Consider a small amount of work dWdW done by a force F⃗\vec{F} over an infinitesimal displacement dr⃗d\vec{r}:

dW=F⃗⋅dr⃗\qquad dW = \vec{F} \cdot d\vec{r}

Now, divide by the time interval dtdt over which this displacement occurs:

P=dWdt=F⃗⋅dr⃗dt\qquad P = \frac{dW}{dt} = \vec{F} \cdot \frac{d\vec{r}}{dt}

Since the time derivative of displacement is the instantaneous velocity (v⃗=dr⃗/dt\vec{v} = d\vec{r}/dt), we get:

P=F⃗⋅v⃗\qquad P = \vec{F} \cdot \vec{v}

This means the instantaneous power delivered to an object is the scalar product of the force acting on it and the object's velocity. It can also be written as P=Fvcos⁡θP = Fv\cos\theta, where θ\theta is the angle between the force and velocity vectors.

Differentiating W = vec Fcdotvec d with respect to time gives this directly. It is the same dot product, one derivative later.
Differentiating W=F⃗⋅d⃗W = \vec F\cdot\vec d with respect to time gives this directly. It is the same dot product, one derivative later.

Why a hill costs more power

P=FvP = Fv makes one comparison very easy. Hold the speed fixed, and the power is decided entirely by the force the engine has to supply.

Speed is the same in both panels, so the whole difference is in the force. P = Fv turns the extra force straight into extra power.
Speed is the same in both panels, so the whole difference is in the force. P=FvP = Fv turns the extra force straight into extra power.

On the level that force only has to beat the resistance. On a slope it also has to lift the car, so the same 20 m/s costs measurably more.

Cruising at a steady speed

A car holding a constant speed on a level road has no change in kinetic energy. Consequently, the net work on it is zero. That does not mean the engine is idle. It means the engine's output is exactly matching what drag and rolling resistance are taking out.

No net force means no change in kinetic energy. The engine's whole output therefore goes into the resistances, at exactly the rate they take it.
No net force means no change in kinetic energy. The engine's whole output therefore goes into the resistances, at exactly the rate they take it.

So the power needed to cruise is P=RvP = Rv, the resistance times the speed. Because RR itself grows with speed, the power needed climbs far faster than the speed does.

Going back from power to work

If power is the derivative of work with respect to time, then work is the integral of power over time. On a graph that means the area, exactly as it did for force against position.

Power is the derivative of work, so work is the area under the power curve. It is the same relationship the force-position graph has, one level up.
Power is the derivative of work, so work is the area under the power curve. It is the same relationship the force-position graph has, one level up.

This is how an engine's output over a whole journey is worked out from a recording of its power. No force and no distance need be known.

A car cruises at a steady 20 m/s against a total resistance of 600 N. What power is the engine delivering?

A sprinter covers 100 m in 10 s. Their average power over the race is 400 W. What can you say about their power at the 2-second mark?

Worked Examples

Illustrative Example: Power of a Car Engine

A 1200 kg car accelerates from rest to a speed of 25 m/s in 8.0 s. Assuming constant acceleration, what is the average power delivered by the engine? (Ignore friction and air resistance).

The engine's work is the car's gain in kinetic energy; dividing by the time it took gives the average power it delivered.
The engine's work is the car's gain in kinetic energy; dividing by the time it took gives the average power it delivered.

Solution:

First, we find the work done by the engine using the Work-Energy Theorem. The net work is done by the engine's force.

1. Initial kinetic energy, Ki=0K_i = 0.

2. Final kinetic energy, Kf=12mvf2=12(1200 kg)(25 m/s)2=12(1200)(625)=375,000 JK_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(1200 \, \text{kg})(25 \, \text{m/s})^2 = \frac{1}{2}(1200)(625) = 375,000 \, \text{J}.

3. The work done by the engine is Wengine=ΔK=Kf−Ki=375,000 JW_{engine} = \Delta K = K_f - K_i = 375,000 \, \text{J}.

Now, we can calculate the average power using the time interval Δt=8.0\Delta t = 8.0 s.

Pavg=WengineΔt=375,000 J8.0 s=46,875 W\qquad P_{avg} = \frac{W_{engine}}{\Delta t} = \frac{375,000 \, \text{J}}{8.0 \, \text{s}} = 46,875 \, \text{W}

The average power delivered by the engine is approximately 46.9 kW, or about 63 hp.

Illustrative Example: Power to Maintain Constant Velocity

A cyclist is riding at a constant velocity of 10 m/s. They are pushing against a total resistive force (air resistance and friction) of 80 N. What is the power output of the cyclist?

Solution:

Since the cyclist is moving at a constant velocity, their acceleration is zero. The net force on them is therefore zero. The forward force applied by the cyclist (FcyclistF_{cyclist}) must be equal in magnitude and opposite in direction to the total resistive force (FresistF_{resist}).

Fcyclist=Fresist=80 N\qquad F_{cyclist} = F_{resist} = 80 \, \text{N}

The force applied by the cyclist is in the same direction as their velocity. We can use the formula P=F⃗⋅v⃗P = \vec{F} \cdot \vec{v}.

P=Fcyclistvcos⁡(0∘)=(80 N)(10 m/s)(1)=800 W\qquad P = F_{cyclist} v \cos(0^\circ) = (80 \, \text{N})(10 \, \text{m/s})(1) = 800 \, \text{W}

The cyclist must maintain a power output of 800 Watts to overcome the resistive forces and maintain a constant speed.