FFR - Power

Original textbook reading.

Knowing the basic energy conversion from kinetic to potential is a nice way of working with energy, and useful in predicting many qualities for various problems in a theoretical sense, but in the real world, we don’t have perfect conversion between the two.

Energy Loss

For a block sliding against a stationary surface, kinetic friction opposes the relative sliding and does negative work on the block. Mechanical energy can become internal thermal energy in the block and surface. Friction need not always do negative work on every possible system; a moving belt can accelerate an object through friction.

When friction takes energy out of an object, it typically is removed in the form of heat. That is, when an object loses energy due to friction, it generates heat. You have probably used this concept already without realizing it. On cold days, you might have rubbed your palms together to get warm. What you are actually doing is converting the kinetic energy of your hands moving into heat energy, which comes from friction between your palms.

Energy Loss to Friction

Now, for the quantitative issue of heat loss, there are a couple different ways. As we say that heat loss comes from frictional forces, we can calculate the work done by those forces as:

For sliding against stationary surfaces, define HL≥0 as the mechanical energy converted to internal thermal energy. With constant friction magnitudes on each segment, HL=Σf_k,iℓ_i (Equation 5.4.1), while signed friction work is W_friction=−HL. The symbol Σ means sum. For the following kinetic-energy sum:

$Σ K E = \sum_{i = 1}^{3} \frac{1}{2} m_{i} {({\vec{v}}_{i})}^{2}$
If we expand this for our example, we get:

$Σ K E = \frac{1}{2} m_{1} {({\vec{v}}_{1})}^{2} + \frac{1}{2} m_{2} {({\vec{v}}_{2})}^{2} + \frac{1}{2} m_{3} {({\vec{v}}_{3})}^{2}$
$Σ K E = \frac{1}{2} \cdot 4 kg \cdot {(5 m / s)}^{2} + \frac{1}{2} \cdot 5 kg \cdot {(3 m / s)}^{2} + \frac{1}{2} \cdot 1 kg \cdot {(6 m / s)}^{2}$

$Σ K E = 90.5 J$

This approach is the more robust definition of heat loss, as it can be used to calculate the loss due to heat if you can simply calculate the frictional forces involved. If you aren’t given that capability, instead you will either calculate heat loss as an energy loss, $H_{L}$ , or you will be given the $H_{L}$ value in the problem.

Now, what that equation gives us is simply a measurement of the sum of energy lost to friction. We still need to incorporate that into the energy balance that we know. This is done by introducing it on one side. As it is lost from the object as it moves from its initial to its final position, we can either subtract it from the initial state’s energy, or add it to the second states energy, as another component that must receive energy. This makes the equation:

$K E_{i} + P E_{g, i} + P E_{s, i} - H_{L} = K E_{f} + P E_{g, f} + P E_{s, f}$
Or

$K E_{i} + P E_{g, i} + P E_{s, i} = K E_{f} + P E_{g, f} + P E_{s, f} + H_{L}$ (Equation 5.4.2)

Whichever you want to use is fine, they both work just as well.

Example Problem 5.4.1

If a $1.5 kg$ ball dropped from $5 m$ hits the ground and bounces up $4.5 m$ , determine how much energy was lost to heat, assuming that was the only source of energy loss in the situation.

Step 1 Pick a datum.

As the two states, are the initial position and the peak of the first bounce, we will pick the datum as the peak of the second bounce. This makes $h_{i} = 0.5 m and h_{2} = 0 m$ .

Step 2 Solve the energy balance

As we have no springs, the energy balance is:
$K E_{i} + P E_{g, i} = K E_{f} + P E_{g, f} + H_{L}$
We have no initial kinetic energy, nor final, as both points are stationary. We can also eliminate $P E_{g, f}$ . as that is at the datum.

$P E_{g, i} = H_{L}$
$1.5 kg \cdot 9.81 m / s^{2} \cdot 0.5 m = H_{L}$

$H_{L} = 7.35 J$

That was a simple problem, but now we will try and solve a more complicated problem, by solving for heat loss through a given frictional force.

Example Problem 5.4.2

If a box is pushed down a ramp, giving it a velocity of $5 m / s$ down the ramp at the position shown in the picture, how far will the box slide past the end of the ramp? The box has a mass of $10 kg$ , and the coefficient of friction between the box and the ramp is $0.2$ . The coefficient of friction between the box and the floor is $0.1$ .

Step 3: Determine the frictional force along the floor

Along the floor, the normal force is calculated the same way, except here the $θ = 0$ , making the normal force equal the gravitational force. This makes friction:

${\vec{F}}_{f r} = m g μ_{k}$

${\vec{F}}_{f r} = 9.81 N$

Step 4: Determine the length the box slides

Here we have an initial state of the box leaving the ramp with some total energy, and the second state of the box without any more energy, being stationary.

$K E_{i} + P E_{i} - H_{L} = K E_{f} + P E_{f}$

As the height does not change, $P E_{i} = P E_{f}$ .

$K E_{i} - H_{L} = K E_{f}$

At the end of the problem, the box has stopped. Therefore $K E_{f} = 0$ .

$K E_{i} = H_{L}$

As there is no potential energy change, we can assume that all the energy left over at the end of the ramp is present in the kinetic energy here:

HL_floor=174.753 J.

And we know that this heat loss is from the work due to friction:

174.753 J=(9.81 N)d.

d=17.814 m≈17.8 m.

Step 1 Determine the friction force along the ramp

The normal force of the box derives from the gravitational force, and is found by the formula:

${\vec{F}}_{N} = {\vec{F}}_{g} \cos θ$
${\vec{F}}_{N} = 9.81 m / s^{2} \cdot 10 kg \cdot \cos 15^{\circ} = 94.8 N$

This makes the frictional force:

${\vec{F}}_{f r} = μ_{k} {\vec{F}}_{N} = 0.2 \cdot 94.8 N = 19.0 N$

Step 2: Determine the energy of the box at the base of the ramp

At the shown height h=2 m on a 15° slope, ramp distance ℓ=2/sin 15°=7.7274 m. Base kinetic energy is K_base=½(10)(5²)+(10)(9.81)(2)−[0.2(10)(9.81)cos 15°]ℓ=174.753 J.

Now wasn’t that a very complicated problem? Heat loss introduces a whole new way of using energy balances to solve problems.

Non-Friction-based Energy Loss

But wait. What if there are other forces involved besides friction? What if we have forces that add to the objects energy, instead of removing it? Again, we have to look at the work the forces do on the object between the states in question. In general, our new equation will be:

$K E_{i} + P E_{i} + W_{a d d e d} = K E_{f} + P E_{f} + H_{L} + W_{l o s t}$
(Equation 5.4.3)

Include each energy transfer exactly once. HereHL denotes mechanical energy converted by friction; W_lost denotes any other outward work transfer. For a general balance one may instead combine them into a single signed nonconservative-work term.

Example Problem 5.4.3

A 5 kg flag is raised vertically 10 m at constant speed with negligible friction. Find the upward force and the work done by that force.

Step 1: Determine the datum

This problem is a very simple one, we can simply choose our datum to be the ground. This makes the $h_{i} = 0 m$ and $h_{f} = 10 m$ .

Step 2: Set up the energy equation

$K E_{i} + P E_{i} + W_{a d d e d} = K E_{f} + P E_{f} + H_{L} + W_{l o s t}$
We can go ahead and eliminate $K E$ , as the flag has the same speed in both states. Its initial and final kinetic energies are equal, so they cancel. We can also eliminate the $H_{L}$ , as friction is neglected. $P E_{i} = 0$ , due to it being level with the datum. The $W_{a d d e d}$ comes from the force added, so we need that. There is no force resisting the flag’s motion, except gravity, which is accounted for in the $P E$ . so $W_{l o s t}$ is eliminated.

This leaves us with:

$W_{a d d e d} = P E_{f}$
$W_{a d d e d} = m g h$
$W_{a d d e d} = 5 kg \cdot 9.81 m / s^{2} \cdot 10 m$
$W_{a d d e d} = 490.5 J$

Step 3: Determine the force

Recall that work is given as:

$W = \vec{F} • \vec{d}$

This allows us to find the force:

$\vec{F} = \frac{\overset{}{W}}{\vec{d}} = \frac{490.5 J}{10 m} = 49.05 N$

As you’ve been dealing with energy, you might be a bit confused on one aspect. The energy that you most commonly deal with, electrical energy is distributed in kWh, or kilowatt-hours. You might be wondering where the time factor comes into play. If you haven’t questioned that yet, well, now you have.

Power

One kilowatt hour is the energy given off by a force providing one kilowatt of power for one hour. It actually converts to joules:

$1 k W h = 3 600 000 J$

This idea of power might be unfamiliar to you, but it actually relates directly to energy, if you haven’t gathered as such already.

Formally, power is the rate at which work changes over time. As you have already seen, work directly relates to energy, so you can also think of power as how fast energy is added or removed from a system. Take note that I said “added or removed.” Although energy is never created or destroyed, it can move from system to system. The system is simply an arbitrary judgment that we make when we set up a problem. For example, if we wanted to look at the energy change of a ball rolling down a hill, that ball is our system. Energy can be added to the ball from outside the system, by either a wind blowing, or a person rolling the ball. That doesn’t mean that the energy is created, it is simply moved into the system.

In equation form, power is written as:

$P = \frac{Δ W}{Δ t}$ (Equation 5.4.4)
The unit of power is the Watt, which is $1 J / s$ .

Example Problem 5.4.4

A 5 kg box is pulled at constant speed 10 m up a 20° ramp in 1 minute. The pull is parallel to the ramp; kinetic friction coefficient is 0.3. Find the average pulling power.

Step 1: Figure out the heat loss due to friction

The normal force in this case will be $46.1 N$ . This makes the frictional force $13.8 N$ .

As the box moves 10m along the ramp, we can find the heat loss by determining the work done by friction.

H_L = −W_friction = fₖℓ = (13.8 N)(10 m) = 138 J.
Notice the sign: friction points against the motion, so its work is negative. The mechanical energy transferred to thermal energy is positive.

Step 2: Set up the energy equation

$K E_{i} + P E_{i} + W_{a d d e d} = K E_{f} + P E_{f} + H_{L} + W_{l o s t}$

Assuming the box is moving at a constant speed, the kinetic energies cancel. If we choose our datum to be at the initial position of the box, that makes :

$h_{1} = 0 m, and h_{2} = 10 \sin 20^{\circ} = 3.42 m$ .
This gives us:

$W_{a d d e d} = 5 kg \cdot 9.81 m / s^{2} \cdot 3.42 m + 138 J$
$W_{a d d e d} = 305.8 J$

Step 3: Determine the power needed.

As work that is added to move from the initial to final state, we just need to divide by time it took to reach the final state:

$P = \frac{Δ W}{Δ t}$
$P = \frac{305.8 J}{60 s}$

$P = 5.10 W$

Pause & recall

Pop Quiz 5.4.1 Power

Question 1 of 5 · 0 correct

AP Topic 3.5 · Read the worked solution

A $500 kg$ elevator is lifted $15 m$ in $1 min$ at a constant speed. How much power does the lifting mechanism exert on the elevator?

Read every full question without answers

Question 1

A $500 kg$ elevator is lifted $15 m$ in $1 min$ at a constant speed. How much power does the lifting mechanism exert on the elevator?

$73.6 kW$
$1.23 kW$
$0.55 kW$
$300 W$

Separate worked answer

Question 2

A $4 kg$ block is sliding along horizontal pavement, which has a $μ_{k} = 0.8$ . If it is sliding at $4 m / s$ at $x = 0$ , how far will it slide before stopping?

$1.02 m$
$4.08 m$
$4.32 m$
$8 m$

Separate worked answer

Question 3

A hamster running in its wheel can run at about $50 cm / s$ . If that hamster has to exert a force of $0.25 N$ to maintain this velocity, how many hamsters are needed to power a $1 MW$ power plant?

$200 000 hamsters$
$500 000 hamsters$
$1 200 000 hamsters$
$8 000 000 hamsters$

Separate worked answer

Question 4

A roller coaster can ascend this $40 m$ high hill, with a $55^{\circ}$ slope in $1$ min at a constant speed. Neglecting friction, how much power does the lifting chain exert on the $1.6 Mg$ train?

Question diagram; use the accompanying question and labels.

$24.4 kW$
$16.5 kW$
$10.5 kW$
$9.44 kW$

Separate worked answer

Question 5

The pictured water lifter has five rising buckets, each containing 5 kg of water, all moving vertically at 1 m/s. Neglect empty-bucket mass, friction, and speed changes. Find the mechanical power required.

$49.05 W$
$98.1 W$
$157.2 W$
$245.3 W$

Separate worked answer

Chapter 5 Equations List

Section 1:

K_f=K_i+W_net. For a constant net force, ½m(v_f²−v_i²)=F_net·Δr; for a varying force use ∫F_net·dr.

Section 3:

U_g=mgh for signed height above a fixed datum. U_s=½kx² for deformation from the unloaded spring length. With only conservative forces doing work, K_i+U_g,i+U_s,i=K_f+U_g,f+U_s,f.

Section 4:

For friction magnitudef_k,i acting over pathlengthℓ_i, HL=Σf_k,iℓ_i≥0. Mechanical energy balance:K_i+U_i+W_added=K_f+U_f+HL +W_lost, counting each transfer once. Average powerP_avg=W/Δt.

Chapter 5 Problems

Work it out

Original problem 1

AP-017: Keep both charges stationary so this option illustrates configuration energy without an ambiguous kinetic-energy increase. Electric potential energy may be negative relative to infinite separation.

Problem text

Which describes a kinetic-energy increase rather than a potential-energy example? A. A watermelon and egg held at height h. B. Two opposite charges held stationary at a fixed separation. C. A stretched spring. D. A runner increasing speed. Explain.

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Work it out

Original problem 2

The result assumes negligible friction, air resistance, and rotational kinetic energy.

Problem text

A snowboarder starts 10 m above the bottom at 3 m/s. Neglect friction and air drag and find the speed at the bottom. Use g = 9.81 m/s².

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Section breakdown

Energy Loss
Friction - the thief of energy
Example 5.4.1 - The Bouncing Ball loses some Energy
Example 5.4.2 - A Box begins its Adventure, only to be stopped by Friction
Friction’s energy-thieving friends
Example 5.4.3 - Raising the Flag
Power
Example 5.4.4 - Using a Ramp to more easily move a Box
Chapter Equations

Chapter Problems

Reviewed physics notes (16)

These corrections keep the original lesson’s purpose while fixing errors in the source edition.

The claim that friction always resists absolute object motion is overgeneralized. — Reference 1, Reference 2
Do not equate positiveheat transfer with a vectorforce multiplied by scalarposition; use pathlength and force magnitude. — Reference 1, Reference 2
Count friction only once: HL covers frictional conversion and W_lost covers other outward work transfers. — Reference 1
The original lacks an acceleration condition; the existing 49.05 N answer assumes constant velocity. — Reference 1, Reference 2
The native solution assumesconstantspeed and ramp-parallelpull; nowexplicit. — Reference 1, Reference 2
Correct equation summary signs andscalarconvention. — Reference 1, Reference 2, Reference 3
The printed gravitational term lacks height, and the source rounds friction before calculating the energy. — Reference 1, Reference 2, Reference 3
Carry the correctedbaseenergy. — Reference 1, Reference 2
Carry the corrected energy consistently. — Reference 1, Reference 2
Retain the rounded final answer with a corrected derivation. — Reference 1, Reference 2
Energy removal per unit time also depends on speed; compare frictional work per unit distance. — Reference 1, Reference 2
The existing 245.3 W key assumes five full rising buckets; state the count and idealizations. — Reference 1
State the net force and constant-force limits in the repeated equation list. — Reference 1, Reference 2
Correct scalar gravitational/spring energies and the mechanical-energy conservation condition in the summary. — Reference 1, Reference 2
Equal nonzero kinetic energies cancel for constant speed; the flag need not be stationary. — Reference 1, Reference 2
Positive thermal energy transfer is −W_friction, not the negative force-displacement dot product. The retained 138 J result is unchanged. — Reference 1, Reference 2

About this edition and its source ↗

Glossary and deeper reading · 9 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Kinetic friction
Kinetic friction acts between surfaces that slide relative to one another. It opposes their relative slipping and is often modeled with magnitude equal to the kinetic-friction coefficient times the normal force. For example, a rapidly moving belt can exert forward kinetic friction on a slower box sliding on it.
Read more: Friction force →
Mechanical energy
Mechanical energy is the sum of the kinetic and potential energies included in a model. It stays constant when transfers to or from other energy stores and across the chosen boundary are absent. For example, a pendulum’s mechanical energy can remain approximately constant while energy shifts between motion and gravitational configuration.
Read more: Mechanical energy in detail →
Heat
Heat is energy transferred between systems because of a temperature difference. It describes a transfer rather than a substance or energy store contained in an object. For example, when a warmer mug transfers energy to a cooler hand, the mug loses internal energy and the hand gains it.
Read more: Heat in detail →
Kinetic energy
Kinetic energy is energy associated with motion relative to a chosen reference frame. For a nonrelativistic translating object it is one half of its mass times speed squared. For example, doubling a cart’s speed quadruples its translational kinetic energy when its mass stays fixed.
Read more: Kinetic energy →
Force
A force describes an interaction that can change an object’s momentum. It is a vector measured in newtons, so both size and direction matter when forces are added. For example, a rope can pull a cart while the floor exerts friction in another direction.
Read more: Free-body diagrams →
Mechanical work
Mechanical work is energy transferred by a force through displacement of its point of application. For a constant force on a particle, it depends on the force component along the particle’s displacement, and it can be positive, negative, or zero. For example, a magnetic force perpendicular to a particle’s displacement does zero work on that particle.
Read more: Work and force →
Power
Power is the rate at which energy is transferred or converted. It is measured in watts, with one watt equal to one joule per second. For example, two motors can lift the same load through the same height but have different average powers if they take different times.
Net force
Net force is the vector sum of all forces acting on the object or system under consideration. For constant mass in an inertial frame, it equals mass times acceleration. For example, equal upward and downward forces give zero net force even though neither force is zero.
Read more: Net force in detail →
Conservative force
A conservative force does work that depends only on the starting and ending configurations. A potential-energy function can describe it, and its work around a closed path is zero. For example, near-Earth gravity does the same work between two heights whether a ball follows a vertical or a curved path.
Read more: Conservative and nonconservative forces →

FFR - Power

Energy Loss

Gallery 5.4.1 Friction

Step 1

Step 2

Energy Loss to Friction

Example Problem 5.4.2

The Floor

Block’s Start

Non-Friction-based Energy Loss

Power

Pop Quiz 5.4.1 Power

Question 1

Question 2

Question 3

Question 4

Question 5

Chapter 5 Equations List

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Kinetic friction

Mechanical energy

Heat

Kinetic energy

Force

Mechanical work

Power

Net force

Conservative force

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