doPhysics.in
Profile

Mathematical Tools & Measurement

  1. 01Basic Essential Mathematics
  2. 02Vectors
  3. 03Differentiation
  4. 04Applications of Differentiation
  5. 05Integration
  6. 06Applications of Integration
  7. 07Physical Quantities and Units
  8. 08Dimensional Formula
  9. 09Dimensional Analysis and Its Applications
  10. 10Experimental Skills

Kinematics

  1. 01Motion in One Dimension
  2. 02Motion in Multiple Dimensions
  3. 03Relative Velocity
  4. 04Circular Motion Kinematics
  5. 05Circular Motion Dynamics

Dynamics

  1. 01Forces and Laws of Motion
  2. 02Laws of Motion
  3. 03Friction
  4. 04Force and Potential Energy
  5. 05Fundamentals of Force
  6. 06Newton's Laws and Free Body Diagrams
  7. 07Applications: Objects in Equilibrium
  8. 08Applications: Objects in Motion
  9. 09Constraint Relations
  10. 10Inertial and Non-Inertial Frames
  11. 11Basics of Friction
  12. 12Applications of Friction

Work, Energy, and Power

  1. 01Conservation of Mechanical Energy
  2. 02Work and the Work-Energy Theorem
  3. 03Work and Kinetic Energy Theorem
  4. 04Energy and its Conservation
Theory/Work, Energy, and Power

Work, Energy, and Power · Chapter 04

Energy and its Conservation

Energy as a stored quantity: potential energy and the conservative forces that permit it, the conservation of mechanical energy and what breaks it, recovering the force from the potential energy curve, reading equilibrium off an energy diagram, and the vertical circle and escape-velocity problems that conservation makes easy.

140 min read · 8 topics

01

Potential Energy

Introduction to Potential Energy

Potential energy (UUU) is the energy stored in a system because of the position or configuration of its parts. Kinetic energy is the energy of motion. Potential energy is stored energy that can later be converted into kinetic energy or used to do work. Potential energy is associated only with a specific class of forces called conservative forces. We will define them formally in the next section. For now, the two most common examples are gravity and the elastic force of a spring.

Suppose a system moves from an initial configuration to a final one. The change in potential energy (ΔU\Delta UΔU) is defined as the negative of the work done (WcW_cWc​) by the conservative force:

ΔU=Uf−Ui=−Wc\qquad \Delta U = U_f - U_i = -W_cΔU=Uf​−Ui​=−Wc​

This definition is crucial. It means:

  • When a conservative force does positive work (e.g. gravity pulling an object down), the potential energy of the system decreases.
  • When a conservative force does negative work (e.g. an external agent lifting an object against gravity), the potential energy of the system increases.

Kinetic energy has a natural zero point: zero speed. Potential energy is defined only by its *change*. We are free to choose a convenient reference configuration and define the potential energy there to be zero (U=0U = 0U=0). The choice of this zero point is arbitrary and does not affect the physics. Only the *difference* in potential energy matters.

What the three have in common is that the energy is in the configuration, not in the motion. Nothing here is moving yet.
What the three have in common is that the energy is in the configuration, not in the motion. Nothing here is moving yet.

If the work done by a conservative force on a system is positive, what happens to the system's potential energy?

What is it that actually holds the energy when a spring is compressed?

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object due to its vertical position in a gravitational field. The position is measured from a chosen reference level. Consider an object of mass mmm near the Earth's surface. There the gravitational force is approximately constant, Fg=mgF_g = mgFg​=mg, acting downwards.

Lift the object vertically from an initial height yiy_iyi​ to a final height yfy_fyf​, through a height h=yf−yih = y_f - y_ih=yf​−yi​. The work done by the gravitational force is:

Wg=Fgdcos⁡(180∘)=(mg)(h)(−1)=−mgh\qquad W_g = F_g d \cos(180^\circ) = (mg)(h)(-1) = -mghWg​=Fg​dcos(180∘)=(mg)(h)(−1)=−mgh

The work is negative because the force of gravity is opposite to the direction of displacement. Now, using the definition of potential energy change:

ΔUg=Uf−Ui=−Wg=−(−mgh)=mgh\qquad \Delta U_g = U_f - U_i = -W_g = -(-mgh) = mghΔUg​=Uf​−Ui​=−Wg​=−(−mgh)=mgh

Choose the reference level yi=0y_i = 0yi​=0, where Ui=0U_i = 0Ui​=0. Then the gravitational potential energy UgU_gUg​ at any height hhh above this level is:

Ug=mgh\qquad U_g = mghUg​=mgh

Lifting at constant speed takes mgh of work against gravity, and that is exactly what the raised mass can give back. The dashed line is where U was chosen to be zero.
Lifting at constant speed takes mghmghmgh of work against gravity, and that is exactly what the raised mass can give back. The dashed line is where UUU was chosen to be zero.

Near the surface the line is straight, so the gradient, which node 05 will show is the force, is the constant mg. Far from the Earth the line curves and the force weakens with it.
Near the surface the line is straight, so the gradient, which node 05 will show is the force, is the constant mgmgmg. Far from the Earth the line curves and the force weakens with it.

Where you measure from is your choice

A height is only ever a height above something. Nothing in the physics picks that something for you. Choose the floor and a book on a table has one potential energy. Choose the table top and the same book has another.

U has no absolute value. The datum is a free choice. Only Delta U appears in any physical prediction, and it is the same in both panels.
UUU has no absolute value. The datum is a free choice. Only ΔU\Delta UΔU appears in any physical prediction, and it is the same in both panels.

Both are correct, and they never disagree about anything that can be measured. Every prediction uses ΔU\Delta UΔU, and the two choices give the same difference. Pick a datum at the start of a problem, write it down, and do not change it halfway.

A ball is held 2 m above a table which is 1 m above the floor. A student takes the floor as the zero of potential energy and a second student takes the table top. On what will the two disagree?

Elastic Potential Energy

Elastic potential energy is the energy stored in an elastic object, such as a spring. It is stored when the object is stretched or compressed from its equilibrium (natural length) position. The force exerted *by* the spring is given by Hooke's Law, Fs=−kxF_s = -kxFs​=−kx, where xxx is the displacement from equilibrium.

Now stretch the spring from an initial displacement xix_ixi​ to a final displacement xfx_fxf​. The work done by the spring force is:

Ws=∫xixfFs(x)dx=∫xixf(−kx)dx=−[12kx2]xixf=−(12kxf2−12kxi2)\qquad W_s = \int_{x_i}^{x_f} F_s(x) dx = \int_{x_i}^{x_f} (-kx) dx = -\left[ \frac{1}{2}kx^2 \right]_{x_i}^{x_f} = -(\frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2)Ws​=∫xi​xf​​Fs​(x)dx=∫xi​xf​​(−kx)dx=−[21​kx2]xi​xf​​=−(21​kxf2​−21​kxi2​)

The change in elastic potential energy is the negative of this work:

ΔUs=Uf−Ui=−Ws=−[−(12kxf2−12kxi2)]=12kxf2−12kxi2\qquad \Delta U_s = U_f - U_i = -W_s = -\left[-(\frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2)\right] = \frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2ΔUs​=Uf​−Ui​=−Ws​=−[−(21​kxf2​−21​kxi2​)]=21​kxf2​−21​kxi2​

Choose the equilibrium position (xi=0x_i = 0xi​=0) as the reference point, where the potential energy is zero (Ui=0U_i = 0Ui​=0). Then the elastic potential energy UsU_sUs​ stored in the spring at a displacement xxx is:

Us=12kx2\qquad U_s = \frac{1}{2}kx^2Us​=21​kx2

This formula holds whether the spring is stretched (x>0x > 0x>0) or compressed (x<0x < 0x<0). The stored energy depends on x2x^2x2.

The work you do stretching the spring is stored in it. The spring gives all of it back when released. That is what makes it a store rather than a loss.
The work you do stretching the spring is stored in it. The spring gives all of it back when released. That is what makes it a store rather than a loss.

The square makes compression and extension equivalent: the spring does not know which way it was deformed, only by how much.
The square makes compression and extension equivalent: the spring does not know which way it was deformed, only by how much.

If you double the distance a spring is stretched, by what factor does its stored potential energy increase?

Whose energy is it?

We say "the ball's potential energy", and it is a convenient shorthand for something slightly different. The energy is stored in the separation between the ball and the Earth. It took both of them to store it.

Neither body owns the energy: it is stored in their separation. We say "the ball's potential energy" as shorthand because the Earth never moves.
Neither body owns the energy: it is stored in their separation. We say "the ball's potential energy" as shorthand because the Earth never moves.

The shorthand is harmless here only because the Earth never visibly moves. It stops being harmless the moment you have two bodies of comparable mass, two stars, or two blocks joined by a spring. Then the potential energy has to be assigned to the pair.

02

Conservative and Non-Conservative Forces

Introduction

In our discussion of potential energy, we introduced the idea of conservative forces. Now we will define them more formally. Forces in nature can be divided into two categories: conservative and non-conservative. This distinction is fundamental to the principle of conservation of energy.

Conservative Forces

A force is considered conservative if the work it does on an object moving between two points is independent of the path taken. It depends only on the initial and final positions.

There is an equivalent definition. A force is conservative if the work it does on an object moving through any closed path is zero. A closed path is one where the starting and ending points are the same.

Properties of a Conservative Force:

  1. Work done is path-independent.
  2. Work done over a closed loop is zero (∮F⃗c⋅dr⃗=0\oint \vec{F}_c \cdot d\vec{r} = 0∮Fc​⋅dr=0).
  3. It can be associated with a potential energy function (F⃗c=−∇U\vec{F}_c = -\nabla UFc​=−∇U).

The relationship ΔU=−Wc\Delta U = -W_cΔU=−Wc​ that we defined earlier is a direct consequence of a force being conservative. The work done depends only on the endpoints. So the change in potential energy, Uf−UiU_f - U_iUf​−Ui​, has one unique value between those two points.

Examples of Conservative Forces

  • Gravitational Force: The work done by gravity when you lift a book depends only on the change in height. It does not matter whether you lift it straight up or take a winding path.
  • Elastic Spring Force: The work done by a spring when it is stretched or compressed depends only on its initial and final displacements from equilibrium.
  • Electrostatic Force: The work done by the electric field on a charge is also path-independent.

Gravity's work depends on the height change and nothing else. That path-independence is the definition of a conservative force.
Gravity's work depends on the height change and nothing else. That path-independence is the definition of a conservative force.

Return to where you started and a conservative force has given back everything it took. Any force that fails this test cannot have a potential energy.
Return to where you started and a conservative force has given back everything it took. Any force that fails this test cannot have a potential energy.

If a force is conservative, is the work it does on an object moving between two points dependent on the path taken?

Why the test matters at all

The test is not a curiosity. Suppose the work between two points depends on the route. Then no number attached to a *point* can have a change equal to that work. A potential energy is exactly such a number.

So everything in the next four nodes rests on passing this one test: U(x)U(x)U(x), energy diagrams, conservation of mechanical energy. A force that fails it does not get a potential energy. Its work has to be carried around explicitly in every calculation.

Non-Conservative Forces

A force is non-conservative if the work it does on an object depends on the path taken between the initial and final points. The work done by a non-conservative force over a closed loop is generally not zero.

Non-conservative forces typically involve dissipative processes like friction or air resistance. These processes convert mechanical energy into other forms, such as heat or sound. Because energy is removed from the mechanical system, you cannot define a potential energy function for these forces.

Examples of Non-Conservative Forces

  • Kinetic Friction: Consider pushing a box across a rough floor from point A to point B and then back to A. The friction force always opposes the motion. On the way to B, friction does negative work. On the way back to A, it *also* does negative work. The total work over the closed path is a negative value, not zero. Furthermore, if you take a longer, roundabout path from A to B, you will do more work against friction.
  • Air Resistance (Drag): Like friction, the drag force always opposes motion. So the work it does depends on the length of the path traveled.
  • Tension in a Rope: While tension can do work, it is generally non-conservative. The work done by tension depends on the specifics of the path and how the tension changes.
  • Applied Forces (Pushes/Pulls): A push or pull from an external agent is typically non-conservative.

Friction charges by the metre, so the two routes cost different amounts. There is no function of position whose change could equal that, which is why friction has no potential energy.
Friction charges by the metre, so the two routes cost different amounts. There is no function of position whose change could equal that, which is why friction has no potential energy.

Which of the following is a classic example of a non-conservative force?

Where the energy actually goes

Nothing is destroyed when friction acts. The mechanical energy, kinetic plus potential, falls. An equal quantity of heat appears in the block and in the floor.

Nothing is destroyed. The energy leaves the mechanical account and turns up as warmth in the block and the floor. No mechanical process can collect it again from there.
Nothing is destroyed. The energy leaves the mechanical account and turns up as warmth in the block and the floor. No mechanical process can collect it again from there.

The reason we call this a loss is practical rather than fundamental. The heat is spread across countless molecules moving in random directions. No mechanical process can gather it back up into the ordered motion of a block.

The whole classification, on one page

Every force this course uses falls on one side or the other of the test. The two questions, does the path matter, and is there a potential energy, always give the same answer.

The two columns always agree, and that is the point: a potential energy exists exactly when the work is path-independent, never otherwise.
The two columns always agree, and that is the point: a potential energy exists exactly when the work is path-independent, never otherwise.

That is not a coincidence. The potential energy is defined as the function whose change is minus the work. Such a function exists exactly when the work does not depend on the route.

Two rows, one test. Gravity gives back on the descent exactly what it took on the climb. Friction charges twice. That is why no potential energy can be written for it.
Two rows, one test. Gravity gives back on the descent exactly what it took on the climb. Friction charges twice. That is why no potential energy can be written for it.

Reading the table across is the quickest form of the test there is. If a force's row totals to zero on every round trip, it is conservative.

A puck slides across a rough table and stops. Its 40 J of kinetic energy has gone. Which statement is right?

A block is carried from the floor up onto a shelf and then straight back down to the floor. Over the whole round trip, what is the total work done by gravity, and by friction?

03

Work Done by Central Forces

What are Central Forces?

A central force F⃗\vec{F}F is one that is always directed along the line connecting a particle to a fixed point. That point is called the center of force. Central forces are a special and important class of forces in physics. Furthermore, the magnitude of the force depends only on the distance rrr from this center, i.e. F=F(r)F = F(r)F=F(r).

Mathematically, a central force can be written as:

F⃗=F(r)r^\vec{F} = F(r)\hat{r}F=F(r)r^

where r^\hat{r}r^ is a unit vector pointing from the center of force towards the particle. The most prominent examples of central forces are the gravitational force between two masses and the electrostatic force between two charges. Both follow an inverse-square law.

A central force has no component along the perpendicular part of the step, so only the change in r can contribute to the work.
A central force has no component along the perpendicular part of the step, so only the change in rrr can contribute to the work.

Central Forces are Conservative

All central forces are conservative. We prove this by showing that the work done by a central force between two points is independent of the path taken.

Consider a particle moving from an initial point AAA to a final point BBB under the influence of a central force. The infinitesimal work done is dW=F⃗⋅ds⃗dW = \vec{F} \cdot d\vec{s}dW=F⋅ds, where ds⃗d\vec{s}ds is the infinitesimal displacement vector. In polar coordinates, this displacement can be written as the sum of a radial component and a tangential component: ds⃗=drr^+rdθθ^d\vec{s} = dr\hat{r} + r d\theta\hat{\theta}ds=drr^+rdθθ^.

We now compute the dot product for the work:

dW=(F(r)r^)⋅(drr^+rdθθ^)dW = (F(r)\hat{r}) \cdot (dr\hat{r} + r d\theta\hat{\theta})dW=(F(r)r^)⋅(drr^+rdθθ^)

Since r^\hat{r}r^ and θ^\hat{\theta}θ^ are orthogonal unit vectors, r^⋅r^=1\hat{r} \cdot \hat{r} = 1r^⋅r^=1 and r^⋅θ^=0\hat{r} \cdot \hat{\theta} = 0r^⋅θ^=0. The expression simplifies:

dW=F(r)drdW = F(r) drdW=F(r)dr

To find the total work done from A to B, we integrate this expression:

WA→B=∫rArBF(r)drW_{A \to B} = \int_{r_A}^{r_B} F(r) drWA→B​=∫rA​rB​​F(r)dr

This result is important. It shows that the work done by a central force depends only on the initial and final radial distances (rAr_ArA​ and rBr_BrB​). It does not depend on the particular path taken between them. Since the work is path-independent, the force is, by definition, conservative. This is why we can define a potential energy function for forces like gravity and the electrostatic force.

Constant r means dr = 0. Every arc segment is free, which is what makes the total depend on the endpoints alone.
Constant rrr means dr=0dr = 0dr=0. Every arc segment is free, which is what makes the total depend on the endpoints alone.

Every path is arcs and radial steps

The argument generalises in one move. Approximate any route by a staircase of arcs at constant radius and steps straight in or out. The arcs cost nothing. The radial steps between the same two radii add to the same total whichever route you took.

Break either route into arcs and radial steps: the arcs contribute nothing and the radial steps between r1 and r2 are the same for both.
Break either route into arcs and radial steps: the arcs contribute nothing and the radial steps between r1r_1r1​ and r2r_2r2​ are the same for both.

So the work depends on r1r_1r1​ and r2r_2r2​ alone. This is the path-independence test of the previous node. Every central force therefore has a potential energy.

A satellite moves from a distance r1r_1r1​ to a distance r2r_2r2​ from the Earth. What does the work done by gravity depend on?

A planet moves along a short arc of its orbit at very nearly constant distance from the Sun. What work does the Sun's gravity do over that arc?

04

Conservation of Mechanical Energy

Introduction

The conservation of energy is one of the most fundamental principles in physics. By combining the Work-Energy Theorem with the idea of conservative forces, we can derive a powerful tool for analyzing motion.

Choosing a System

First, we must define our system. The choice of system is crucial as it determines which forces are internal and which are external. Energy is conserved within a closed, isolated system. In such a system, no external force does work and no internal non-conservative force is present.

Total Mechanical Energy

The total mechanical energy (EEE) of a system is the sum of its kinetic energy (KKK) and its total potential energy (UUU). The potential energy UUU is the sum of all types of potential energy within the system (e.g. gravitational, elastic).

E=K+U\qquad E = K + UE=K+U

The Generalized Work-Energy Principle

Consider the net work done on an object, split into work done by conservative forces (WcW_cWc​) and work done by non-conservative forces (WncW_{nc}Wnc​):

Wnet=Wc+Wnc\qquad W_{net} = W_c + W_{nc}Wnet​=Wc​+Wnc​

From the Work-Energy Theorem, Wnet=ΔKW_{net} = \Delta KWnet​=ΔK. From the definition of potential energy, Wc=−ΔUW_c = -\Delta UWc​=−ΔU. Substituting these gives:

ΔK=−ΔU+Wnc\qquad \Delta K = -\Delta U + W_{nc}ΔK=−ΔU+Wnc​

Rearranging this yields the generalized work-energy principle:

ΔK+ΔU=WncorΔE=Wnc\qquad \Delta K + \Delta U = W_{nc} \quad \text{or} \quad \Delta E = W_{nc}ΔK+ΔU=Wnc​orΔE=Wnc​

This is a more complete statement about energy. It says that the change in the total mechanical energy of a system is equal to the work done by all non-conservative forces. If WncW_{nc}Wnc​ is negative (e.g. from friction), the system's total mechanical energy decreases.

Steps 1 and 2 are the ones readers skip, and they are where the errors come from. A force gets counted twice, or two heights are measured from different places.
Steps 1 and 2 are the ones readers skip, and they are where the errors come from. A force gets counted twice, or two heights are measured from different places.

The Principle of Conservation of Mechanical Energy

A special and very important case arises when the work done by non-conservative forces is zero (Wnc=0W_{nc} = 0Wnc​=0). This happens if there are no non-conservative forces, like friction or air resistance. It also happens if they act but do no work, for example a normal force perpendicular to the motion.

In this case, ΔE=0\Delta E = 0ΔE=0. This means the total mechanical energy of the system remains constant.

Ef=EiorKf+Uf=Ki+Ui\qquad E_f = E_i \quad \text{or} \quad K_f + U_f = K_i + U_iEf​=Ei​orKf​+Uf​=Ki​+Ui​

This is the Principle of Conservation of Mechanical Energy. It is a powerful problem-solving tool. It relates the speeds and positions of an object at two different points without analyzing the motion in between. The only condition is that conservative forces alone are doing work.

The split between the two colours changes at every point on the track; the total height of the bar does not. That constant total is the mechanical energy.
The split between the two colours changes at every point on the track; the total height of the bar does not. That constant total is the mechanical energy.

A block slides down a smooth curved track. Which condition actually guarantees that its mechanical energy is conserved?

The same total, instant by instant

A falling ball is the smallest example that shows the whole idea. The potential bar shrinks and the kinetic bar grows. Their sum is the same number at every height.

At every instant K + U is the same number. That is the whole content of conservation, and it is why the final speed follows without a single equation of motion.
At every instant K+UK + UK+U is the same number. That is the whole content of conservation, and it is why the final speed follows without a single equation of motion.

A pendulum is the same statement made periodic. The bob rises to exactly the height it was released from, on both sides. The tension is always perpendicular to the motion, so it does no work at all.

The bob rises to exactly the height it was released from, on both sides. This holds as long as the only forces are gravity and a tension that does no work.
The bob rises to exactly the height it was released from, on both sides. This holds as long as the only forces are gravity and a tension that does no work.

Why the shape of the track drops out

The most useful thing about conservation is what it does not need to know. It relates two instants and says nothing about the route between them. So a complicated frictionless track is no harder than a simple one.

The normal force does no work whatever shape the track is, so only the drop in height survives into the answer.
The normal force does no work whatever shape the track is, so only the drop in height survives into the answer.

All three ramps deliver the same speed at the bottom. The normal force does no work whatever shape the surface has, so only the drop in height survives into the answer. Solving any of these with forces would mean handling an acceleration that changes direction all the way down.

Three frictionless slides drop from the same height to the same floor: one straight, one steep-then-shallow, one shallow-then-steep. Compare the speed at the bottom and the time taken.

And when it fails

Put friction on the same ramp and the total bar is shorter at the bottom than it was at the top. Mechanical energy is not conserved. Total energy still is, and the difference is sitting in the surfaces as heat.

Total energy is still conserved; mechanical energy is not. The heat slice is exactly fk times the distance slid.
Total energy is still conserved; mechanical energy is not. The heat slice is exactly fkf_kfk​ times the distance slid.

The bookkeeping is exact. The missing mechanical energy equals fkf_kfk​ times the distance slid. That turns the generalized principle above into a practical method. Write down Etop−Ebottom=fkdE_{\text{top}} - E_{\text{bottom}} = f_k dEtop​−Ebottom​=fk​d and solve.

Draw the boundary before you do the bookkeeping

Gravity may appear as an external force doing work, or as a potential energy stored inside the system. The choice is not a fact about gravity. It is a consequence of where you drew the system boundary.

Potential energy only exists once both interacting bodies are inside the boundary. Draw the boundary first; the bookkeeping follows from it.
Potential energy only exists once both interacting bodies are inside the boundary. Draw the boundary first; the bookkeeping follows from it.

Take the ball alone and the Earth is outside. Consequently, gravity is external and does work on the system. Take the ball and the Earth together and there is no external force at all. The same energy appears as UUU inside. Both accounts give the same answer. Mixing them double-counts.

Worked Examples

Illustrative Example: Falling Ball

A 0.5 kg ball is dropped from rest from a height of 10 m. Ignoring air resistance, what is its speed just before it hits the ground?

The mass cancels, so every object dropped from the same height arrives at the same speed. This is Galileo's result, obtained here without mentioning acceleration.
The mass cancels, so every object dropped from the same height arrives at the same speed. This is Galileo's result, obtained here without mentioning acceleration.

Solution:

The only force doing work is gravity, which is a conservative force. Therefore, mechanical energy is conserved. Take the ground as the reference level (y=0y=0y=0. Consequently, U=0U=0U=0).

Initial State (at height h):

  • hi=10h_i = 10hi​=10 m
  • vi=0v_i = 0vi​=0 m/s (dropped from rest)
  • Ki=12mvi2=0K_i = \frac{1}{2}mv_i^2 = 0Ki​=21​mvi2​=0
  • Ui=mghi=(0.5 kg)(9.8 m/s2)(10 m)=49 JU_i = mgh_i = (0.5 \, \text{kg})(9.8 \, \text{m/s}^2)(10 \, \text{m}) = 49 \, \text{J}Ui​=mghi​=(0.5kg)(9.8m/s2)(10m)=49J
  • Ei=Ki+Ui=0+49=49 JE_i = K_i + U_i = 0 + 49 = 49 \, \text{J}Ei​=Ki​+Ui​=0+49=49J

Final State (just before hitting the ground):

  • hf=0h_f = 0hf​=0 m
  • vf=?v_f = ?vf​=?
  • Kf=12mvf2=12(0.5)vf2=0.25vf2K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(0.5)v_f^2 = 0.25 v_f^2Kf​=21​mvf2​=21​(0.5)vf2​=0.25vf2​
  • Uf=mghf=0U_f = mgh_f = 0Uf​=mghf​=0
  • Ef=Kf+Uf=0.25vf2E_f = K_f + U_f = 0.25 v_f^2Ef​=Kf​+Uf​=0.25vf2​

Apply Conservation of Energy:

Ef=Ei\qquad E_f = E_iEf​=Ei​

0.25vf2=49\qquad 0.25 v_f^2 = 490.25vf2​=49

vf2=490.25=196\qquad v_f^2 = \frac{49}{0.25} = 196vf2​=0.2549​=196

vf=196=14 m/s\qquad v_f = \sqrt{196} = 14 \, \text{m/s}vf​=196​=14m/s

The ball's speed just before hitting the ground is 14 m/s. We solved this without using the kinematics equations for constant acceleration. That shows the advantage of the energy conservation approach.

Illustrative Example: Block on a Frictionless Ramp

A 2 kg block starts from rest at the top of a frictionless ramp of height 5 m. What is its speed at the bottom of the ramp?

The angle never enters the energy equation. Only the vertical drop does, so the answer is the same sqrt2gh as a straight fall.
The angle never enters the energy equation. Only the vertical drop does, so the answer is the same 2gh\sqrt{2gh}2gh​ as a straight fall.

Solution:

The forces acting on the block are gravity and the normal force. The normal force is always perpendicular to the displacement along the ramp, so it does no work. Gravity is a conservative force. Thus, mechanical energy is conserved. Let the bottom of the ramp be the reference level (y=0y=0y=0).

Initial State (top of ramp):

  • hi=5h_i = 5hi​=5 m, vi=0v_i = 0vi​=0 m/s
  • Ei=Ki+Ui=0+mghi=(2 kg)(9.8 m/s2)(5 m)=98 JE_i = K_i + U_i = 0 + mgh_i = (2 \, \text{kg})(9.8 \, \text{m/s}^2)(5 \, \text{m}) = 98 \, \text{J}Ei​=Ki​+Ui​=0+mghi​=(2kg)(9.8m/s2)(5m)=98J

Final State (bottom of ramp):

  • hf=0h_f = 0hf​=0 m, vf=?v_f = ?vf​=?
  • Ef=Kf+Uf=12mvf2+0=12(2)vf2=vf2E_f = K_f + U_f = \frac{1}{2}mv_f^2 + 0 = \frac{1}{2}(2)v_f^2 = v_f^2Ef​=Kf​+Uf​=21​mvf2​+0=21​(2)vf2​=vf2​

Apply Conservation of Energy:

Ef=Ei\qquad E_f = E_iEf​=Ei​

vf2=98\qquad v_f^2 = 98vf2​=98

vf=98≈9.9 m/s\qquad v_f = \sqrt{98} \approx 9.9 \, \text{m/s}vf​=98​≈9.9m/s

The block's speed at the bottom is approximately 9.9 m/s. Note that the result does not depend on the angle of the ramp, only the height.

A 2 kg block slides down a rough slope. It arrives at the bottom with 30 J of kinetic energy, having lost 50 J of potential energy. How much energy went into heat, and is energy conserved?

05

Relation of Potential Energy and Force

Introduction

We have defined the change in potential energy as the negative of the work done by a conservative force: ΔU=−Wc\Delta U = -W_cΔU=−Wc​. This integral relationship can be inverted to express the force in terms of the potential energy. This provides a powerful way to find the force if you know the potential energy landscape of a system.

Force from Potential Energy in One Dimension

We start with the one-dimensional case. The work done by a conservative force Fx(x)F_x(x)Fx​(x) as a particle moves a small distance Δx\Delta xΔx is approximately Wc≈Fx(x)ΔxW_c \approx F_x(x) \Delta xWc​≈Fx​(x)Δx. The corresponding change in potential energy is:

ΔU≈−Wc≈−Fx(x)Δx\Delta U \approx -W_c \approx -F_x(x) \Delta xΔU≈−Wc​≈−Fx​(x)Δx

Rearranging this, we get:

Fx(x)≈−ΔUΔxF_x(x) \approx -\frac{\Delta U}{\Delta x}Fx​(x)≈−ΔxΔU​

To find the exact force at a point, we take the limit as Δx\Delta xΔx approaches zero. This limit is the definition of the derivative:

Fx(x)=−dU(x)dxF_x(x) = -\frac{dU(x)}{dx}Fx​(x)=−dxdU(x)​

This is the crucial result. The conservative force on an object is the negative of the slope of its potential energy curve with respect to position. A steep slope on a U vs. x graph implies a large force. A flat slope (zero slope) implies zero force.

The force always points the way the potential energy decreases: downhill on this curve, exactly as a ball rolls downhill on a real one.
The force always points the way the potential energy decreases: downhill on this curve, exactly as a ball rolls downhill on a real one.

Reading the sign off the slope

The minus sign is the whole content of the formula. It is easiest to trust once you have read it off three cases.

A flat stretch of U is a force-free region, which is the same statement as "constant potential energy means no force" made visible.
A flat stretch of UUU is a force-free region, which is the same statement as "constant potential energy means no force" made visible.

Uphill to the right means the force points left. Downhill to the right means it points right. Flat means no force. In every case the force points the way the potential energy falls. A ball on a real hillside does the same.

The potential energy U(x) of a particle is constant over a region of space. What is the force Fx on the particle in that region?

Force from Potential Energy in Three Dimensions

In three dimensions, the potential energy UUU can be a function of x, y, and z. The force vector F⃗\vec{F}F has three components. Each component is the negative partial derivative of the potential energy with respect to the corresponding coordinate:

Fx=−∂U∂x,Fy=−∂U∂y,Fz=−∂U∂zF_x = -\frac{\partial U}{\partial x}, \quad F_y = -\frac{\partial U}{\partial y}, \quad F_z = -\frac{\partial U}{\partial z}Fx​=−∂x∂U​,Fy​=−∂y∂U​,Fz​=−∂z∂U​

This can be expressed more compactly using the gradient operator (∇\nabla∇):

F⃗=−∇U=−(∂U∂xi^+∂U∂yj^+∂U∂zk^)\vec{F} = -\nabla U = -\left( \frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k} \right)F=−∇U=−(∂x∂U​i^+∂y∂U​j^​+∂z∂U​k^)

The gradient vector ∇U\nabla U∇U points in the direction in which the potential energy UUU increases most rapidly. The negative sign means that the conservative force F⃗\vec{F}F always points in the direction of decreasing potential energy. Particles are pushed 'downhill' on the potential energy landscape.

The force is at right angles to the contours, because moving along a contour changes no potential energy and therefore costs no work.
The force is at right angles to the contours, because moving along a contour changes no potential energy and therefore costs no work.

Worked Example

Illustrative Example: Spring Force from Elastic Potential Energy

The potential energy stored in an ideal spring is given by U(x)=12kx2U(x) = \frac{1}{2}kx^2U(x)=21​kx2, where xxx is the displacement from equilibrium. Find the force exerted by the spring.

Solution:

We use the one-dimensional relationship Fx=−dU/dxF_x = -dU/dxFx​=−dU/dx.

Fx=−ddx(12kx2)F_x = -\frac{d}{dx} \left( \frac{1}{2}kx^2 \right)Fx​=−dxd​(21​kx2)

Taking the derivative with respect to x gives:

Fx=−12k(2x)=−kxF_x = -\frac{1}{2}k(2x) = -kxFx​=−21​k(2x)=−kx

This is precisely Hooke's Law for the restoring force of a spring. The negative sign indicates that it is a restoring force. It always points back towards the equilibrium position (x=0), which is the position of minimum potential energy.

Differentiating tfrac12kx2 returns -kx: Hooke's law and the spring's stored energy are two ways of writing one fact.
Differentiating 12kx2\tfrac{1}{2}kx^221​kx2 returns −kx-kx−kx: Hooke's law and the spring's stored energy are two ways of writing one fact.

Where U is steepest the force is largest; where U bottoms out the force crosses zero. Differentiating the top curve gives the bottom one.
Where UUU is steepest the force is largest; where UUU bottoms out the force crosses zero. Differentiating the top curve gives the bottom one.

A particle sits at a point where the potential energy curve is at a local minimum. What force acts on it there, and what happens if it is nudged?

06

Energy Diagrams and Equilibrium

Introduction to Energy Diagrams

A powerful tool for visualizing and analyzing the motion of a system is the potential energy diagram, or energy diagram. This is a graph of the system's potential energy UUU as a function of the position xxx of one of its particles. Since Fx=−dU/dxF_x = -dU/dxFx​=−dU/dx, we can infer the force on the particle by looking at the negative of the slope of the curve.

Interpreting Energy Diagrams

The total mechanical energy E=K+UE = K + UE=K+U of the system is represented by a horizontal line on the energy diagram. You may picture the potential energy curve as a frictionless roller coaster track. Kinetic energy K=E−UK = E - UK=E−U can never be negative (K≥0K \ge 0K≥0). So the particle can move only in regions where the total energy line is above the potential energy curve (E≥UE \ge UE≥U).

  • Kinetic Energy (K): At each position, the kinetic energy is the vertical distance from the curve U(x)U(x)U(x) up to the total energy line EEE.
  • Force (F_x): The force is the negative of the slope of the U(x) curve. Where the slope is negative (downhill), the force is positive (to the right). Where the slope is positive (uphill), the force is negative (to the left).
  • Turning Points: These are the points where the total energy line intersects the potential energy curve (E=UE = UE=U). At these points, the kinetic energy is momentarily zero (K=0K=0K=0). The particle reverses its direction of motion.
  • Classically Forbidden Regions: Regions where the potential energy curve is above the total energy line (U>EU > EU>E) are inaccessible to the particle. Entering them would require negative kinetic energy.

The motion is trapped between x1 and x2. Beyond them the kinetic energy would have to be negative. That is why the particle turns around there.
The motion is trapped between x1x_1x1​ and x2x_2x2​. Beyond them the kinetic energy would have to be negative. That is why the particle turns around there.

Move the line, change the motion

The curve is a property of the forces. The horizontal line is a property of how much energy this particular particle has. Moving it changes the answer completely.

Whether the motion is trapped is not a property of the curve alone. It is a question about where the energy line sits on it.
Whether the motion is trapped is not a property of the curve alone. It is a question about where the energy line sits on it.

Low down, the line cuts the curve twice and the particle is trapped between those two crossings. High enough, one of the crossings disappears. The particle then escapes to the right and never comes back.

The right-hand valley is lower, and the particle still cannot get to it: energy decides what is reachable, not what is lowest.
The right-hand valley is lower, and the particle still cannot get to it: energy decides what is reachable, not what is lowest.

A landscape with two valleys makes the point sharply. The right-hand valley is deeper, and a particle in the left one still cannot reach it. What is reachable is decided by the energy, not by what is lowest.

The figure below makes the energy line a control. Choose a value of E and press Play. The turning points move to wherever the curve crosses the chosen line.

Choose an energy level and press Play. A larger total energy pushes the two turning points apart, and the gap between line and curve is the kinetic energy.
Choose an energy level and press Play. A larger total energy pushes the two turning points apart, and the gap between line and curve is the kinetic energy.

Equilibrium Points

Equilibrium occurs at positions where the net force on the particle is zero. Since Fx=−dU/dxF_x = -dU/dxFx​=−dU/dx, this corresponds to points where the slope of the potential energy curve is zero (i.e. at local minima or maxima).

  • Stable Equilibrium: Occurs at a local minimum of the potential energy curve. If the particle is displaced slightly from this point, the force will be a restoring force, pushing it back towards the minimum. Think of a marble at the bottom of a bowl.
  • Unstable Equilibrium: Occurs at a local maximum of the potential energy curve. If the particle is displaced slightly, the force will push it further away from the equilibrium point. Think of a marble balanced on top of a hill.
  • Neutral Equilibrium: Occurs in a region where the potential energy is constant (a flat line). If the particle is displaced, it will remain at the new position with no force acting on it. Think of a marble on a flat, level table.

All three have dU/dx = 0, so all three are equilibria. The second derivative decides whether a nudge is undone, amplified, or simply kept.
All three have dU/dx=0dU/dx = 0dU/dx=0, so all three are equilibria. The second derivative decides whether a nudge is undone, amplified, or simply kept.

On an energy diagram, where would you find a point of stable equilibrium?

On an energy diagram the total-energy line crosses the potential energy curve at x=x1x = x_1x=x1​. What is happening to the particle there?

A marble is balanced exactly on the crest of a potential energy hill. Is it in equilibrium?

07

Application: The Dynamics of Vertical Circular Motion

Introduction

An object moving in a vertical circle is a classic problem. Examples are a roller coaster car in a loop-the-loop and a ball swung on a string. The problem uses both energy methods and Newton's laws. Energy conservation relates the speeds at different heights. Newton's second law for circular motion (∑Fradial=mv2/r\sum F_{radial} = mv^2/r∑Fradial​=mv2/r) gives the forces involved.

Analyzing the Motion

Consider a small object of mass mmm moving in a vertical circle of radius rrr. Set the potential energy reference U=0U=0U=0 at the bottom of the circle. At any angle θ\thetaθ from the bottom, the height of the object is h=r(1−cos⁡θ)h = r(1 - \cos\theta)h=r(1−cosθ).

Forces at the Top and Bottom

We now analyze the two most important points: the top and the bottom of the loop.

At the Bottom (θ=0∘\theta = 0^\circθ=0∘, h=0):

  • Forces: The tension TbotT_{bot}Tbot​ (or normal force) acts upwards. Gravity mgmgmg acts downwards.
  • Newton's Second Law: The net force provides the centripetal acceleration (upwards).
    Tbot−mg=mvbot2r  ⟹  Tbot=mg+mvbot2rT_{bot} - mg = \frac{mv_{bot}^2}{r} \implies T_{bot} = mg + \frac{mv_{bot}^2}{r}Tbot​−mg=rmvbot2​​⟹Tbot​=mg+rmvbot2​​

At the Top (θ=180∘\theta = 180^\circθ=180∘, h=2r):

  • Forces: Both tension TtopT_{top}Ttop​ (or normal force) and gravity mgmgmg act downwards.
  • Newton's Second Law: The net force provides the centripetal acceleration (downwards).
    Ttop+mg=mvtop2r  ⟹  Ttop=mvtop2r−mgT_{top} + mg = \frac{mv_{top}^2}{r} \implies T_{top} = \frac{mv_{top}^2}{r} - mgTtop​+mg=rmvtop2​​⟹Ttop​=rmvtop2​​−mg

Only the component of the weight along the string competes with the tension. The other component is what speeds the bob up and slows it down.
Only the component of the weight along the string competes with the tension. The other component is what speeds the bob up and slows it down.

Same circle, same string, opposite bookkeeping: gravity helps supply the centripetal force at the top and works against it at the bottom.
Same circle, same string, opposite bookkeeping: gravity helps supply the centripetal force at the top and works against it at the bottom.

The Critical Speed

We now find the minimum speed the object must have at the top of the loop to stay on the circular path. If the object is a block on a track, this is the speed at which it is about to lose contact. If it is a ball on a string, this is the speed at which the string is about to go slack. In both cases, the condition is that the tension or normal force becomes zero (Ttop=0T_{top} = 0Ttop​=0 or Ntop=0N_{top} = 0Ntop​=0).

Setting Ttop=0T_{top} = 0Ttop​=0 in our equation for the top:

0+mg=mvcrit2r0 + mg = \frac{mv_{crit}^2}{r}0+mg=rmvcrit2​​

Solving for this minimum or 'critical' speed at the top gives:

vcrit,top=grv_{crit, top} = \sqrt{gr}vcrit,top​=gr​

If the speed at the top is less than gr\sqrt{gr}gr​, the object will fall off the path and become a projectile.

Below sqrtgr the string would need to push, and a string cannot push, so the bob leaves the circular path instead.
Below gr\sqrt{gr}gr​ the string would need to push, and a string cannot push, so the bob leaves the circular path instead.

The figure below tests the condition at three launch speeds. Watch the tension gauge as the bob climbs.

Three launch speeds, one test. Above the critical speed the string stays taut all round. Below it the tension reaches zero on the way up, and the bob falls freely.
Three launch speeds, one test. Above the critical speed the string stays taut all round. Below it the tension reaches zero on the way up, and the bob falls freely.

At the very top of a roller coaster loop, the centripetal force is provided by what combination of forces?

Where in a vertical circle is the string's tension greatest, for a bob going round at a steady rate of turning?

Paying for the height

The critical condition is about forces at one instant. Getting to that instant at all is an energy question. The two have to be used together.

Getting round needs enough kinetic energy at the bottom to buy 2mgr of height and still have tfrac12mgr left at the top.
Getting round needs enough kinetic energy at the bottom to buy 2mgr2mgr2mgr of height and still have 12mgr\tfrac{1}{2}mgr21​mgr left at the top.

Rising to the top costs 2mgr2mgr2mgr out of the kinetic energy at the bottom. What is left has to be at least 12m(gr)\tfrac{1}{2}m(gr)21​m(gr). Adding those gives the familiar u2≥5gru^2 \ge 5gru2≥5gr at the lowest point.

Worked Example

Illustrative Example: Minimum Height for a Roller Coaster

A roller coaster car starts from rest at a height hhh. What is the minimum height hhh needed to complete a vertical loop of radius rrr without losing contact with the track?

Solution:

We need to find the height hhh that results in the car having exactly the critical speed vtop=grv_{top} = \sqrt{gr}vtop​=gr​ at the top of the loop. We use conservation of mechanical energy. It relates the initial state (at height h) to the state at the top of the loop (at height 2r).

  1. Initial State:
    Ei=Ki+Ui=0+mghE_i = K_i + U_i = 0 + mghEi​=Ki​+Ui​=0+mgh
  2. Final State (at the top of the loop):
    Ef=Kf+Uf=12mvtop2+mg(2r)E_f = K_f + U_f = \frac{1}{2}mv_{top}^2 + mg(2r)Ef​=Kf​+Uf​=21​mvtop2​+mg(2r)

Now, substitute the critical speed condition vtop2=grv_{top}^2 = grvtop2​=gr into the final energy equation:

Ef=12m(gr)+2mgr=52mgrE_f = \frac{1}{2}m(gr) + 2mgr = \frac{5}{2}mgrEf​=21​m(gr)+2mgr=25​mgr

By conservation of energy, Ei=EfE_i = E_fEi​=Ef​:

mgh=52mgrmgh = \frac{5}{2}mgrmgh=25​mgr

The mass mmm and ggg cancel out, leaving the minimum height:

hmin=52rh_{min} = \frac{5}{2}rhmin​=25​r

To complete the loop, the roller coaster must start from a height of at least 2.5 times the loop's radius.

Two ideas, one line each: conservation gives the speed at the top, and the critical condition says what that speed has to be.
Two ideas, one line each: conservation gives the speed at the top, and the critical condition says what that speed has to be.

A ball on a light rigid rod, rather than a string, is swung in a vertical circle. Does it still need a speed of at least gr\sqrt{gr}gr​ at the top?

08

Further Applications of Energy Conservation

Introduction

We now apply the laws of energy conservation to a range of important problems. These examples show the usefulness of thinking in terms of energy.

Escape Velocity

The escape velocity is the minimum initial speed an object needs to escape the gravitational pull of a planet and travel infinitely far away. At this minimum speed, the object has exactly zero total mechanical energy and arrives at an infinite distance with zero speed.

For this problem, we must use the universal law of gravitation. Its potential energy is U(r)=−GMm/rU(r) = -GMm/rU(r)=−GMm/r, with the zero of potential energy chosen at r=∞r = \inftyr=∞.

  1. Initial State (on the planet's surface):
    Radius ri=Rplanetr_i = R_{planet}ri​=Rplanet​, initial speed vi=vescv_i = v_{esc}vi​=vesc​.
    Ei=Ki+Ui=12mvesc2−GMmRplanetE_i = K_i + U_i = \frac{1}{2}mv_{esc}^2 - \frac{GMm}{R_{planet}}Ei​=Ki​+Ui​=21​mvesc2​−Rplanet​GMm​
  2. Final State (infinitely far away):
    Distance rf=∞r_f = \inftyrf​=∞, final speed vf=0v_f = 0vf​=0.
    Ef=Kf+Uf=0+0=0E_f = K_f + U_f = 0 + 0 = 0Ef​=Kf​+Uf​=0+0=0

By conservation of energy, Ei=EfE_i = E_fEi​=Ef​:

12mvesc2−GMmRplanet=0\frac{1}{2}mv_{esc}^2 - \frac{GMm}{R_{planet}} = 021​mvesc2​−Rplanet​GMm​=0

Solving for the escape velocity, vescv_{esc}vesc​:

vesc=2GMRplanetv_{esc} = \sqrt{\frac{2GM}{R_{planet}}}vesc​=Rplanet​2GM​​

For Earth, this speed is approximately 11.2 km/s.

Escape means arriving at infinity with nothing left over, so the launch kinetic energy has to equal the whole depth of the potential well.
Escape means arriving at infinity with nothing left over, so the launch kinetic energy has to equal the whole depth of the potential well.

The escaping mass appears on both sides of tfrac12mve2 = GMm/R and cancels. A heavier craft needs more energy, and exactly the same speed.
The escaping mass appears on both sides of 12mve2=GMm/R\tfrac{1}{2}mv_e^2 = GMm/R21​mve2​=GMm/R and cancels. A heavier craft needs more energy, and exactly the same speed.

Does the escape velocity from a planet depend on the mass of the escaping object?

A probe is launched from Earth with a total energy that is negative. What happens to it?

The well, drawn

Taking U=0U = 0U=0 at infinity is the only choice that makes "escaped" mean something. It puts every bound body at a negative potential energy. The curve climbs towards zero as the distance grows.

A negative total energy means the particle is bound and has a furthest distance it can reach. Zero is the dividing line, and it fixes ve = sqrt2GM/R.
A negative total energy means the particle is bound and has a furthest distance it can reach. Zero is the dividing line, and it fixes ve=2GM/Rv_e = \sqrt{2GM/R}ve​=2GM/R​.

A total energy below zero means the line meets the curve at some finite distance. That is the furthest the body can get before falling back. A total energy of exactly zero is the dividing case. Setting it gives ve=2GM/Rv_e = \sqrt{2GM/R}ve​=2GM/R​ at once.

Bungee Jumping

A bungee jump converts energy between three forms: gravitational potential energy, elastic potential energy, and kinetic energy. Let the unstretched length of the bungee cord be LLL and its spring constant be kkk. We set the zero of gravitational potential energy (Ug=0U_g=0Ug​=0) at the lowest point of the jump.

We now analyze the point where the jumper has fallen a total distance HHH and momentarily comes to rest at the bottom.

  1. Initial State (at the start of the jump):
    Height yi=Hy_i = Hyi​=H, speed vi=0v_i = 0vi​=0. The cord is slack.
    Ei=Ki+Ug,i+Us,i=0+mgH+0E_i = K_i + U_{g,i} + U_{s,i} = 0 + mgH + 0Ei​=Ki​+Ug,i​+Us,i​=0+mgH+0
  2. Final State (at the lowest point):
    Height yf=0y_f = 0yf​=0, speed vf=0v_f = 0vf​=0. The cord is stretched by an amount x=H−Lx = H - Lx=H−L.
    Ef=Kf+Ug,f+Us,f=0+0+12k(H−L)2E_f = K_f + U_{g,f} + U_{s,f} = 0 + 0 + \frac{1}{2}k(H-L)^2Ef​=Kf​+Ug,f​+Us,f​=0+0+21​k(H−L)2

By conservation of energy, Ei=EfE_i = E_fEi​=Ef​:

mgH=12k(H−L)2mgH = \frac{1}{2}k(H-L)^2mgH=21​k(H−L)2

This equation relates the total drop distance HHH to the jumper's mass and the cord's properties. It is a quadratic equation that can be solved for HHH to determine how far the jumper will fall.

The fastest point is not the lowest one. It is where the cord's pull first balances the weight. After that the jumper is still descending, but slowing.
The fastest point is not the lowest one. It is where the cord's pull first balances the weight. After that the jumper is still descending, but slowing.

Three stores, one constant total. At the lowest point everything the jumper started with is sitting in the stretched cord.
Three stores, one constant total. At the lowest point everything the jumper started with is sitting in the stretched cord.

At which point of a bungee jump is the jumper moving fastest?

PreviousWork and Kinetic Energy Theorem
doPhysics
TheoryPricingAboutPrivacy PolicyTerms and Conditions

© 2026 doPhysics. All rights reserved.

Engineered by Ankit Shukla